Integrated Math III Honors

Unit 1 7 topics

Sequences & Series

Finite Geometric Series.

1.1 Finite Geometric Series

1.1.1
The Sum of a Finite Geometric Series
A geometric series adds terms that each get multiplied by the same ratio $r$ . With first term $a$ , ratio $r \ne 1$ , and $n$ terms, use the sum formula.
e.g. $2 + 6 + 18 + 54$ : $a = 2$ , $r = 3$ , $n = 4$ , so $S = \frac{2(1 - 81)}{1 - 3} = 80$
$\displaystyle S_n = \frac{a(1 - r^n)}{1 - r}$
1.1.2
The Sum of the First N Terms of a Geometric Series
Identify the first term $a_1$ , the ratio $r$ , and the number of terms $n$ , then substitute. If $r = 1$ , every term is the same and the sum is $n a_1$ .
e.g. First 6 terms of $1, 2, 4, \dots$ : $S_6 = \frac{1(1 - 64)}{1 - 2} = 63$
$\displaystyle S_n = \frac{a_1(1 - r^n)}{1 - r}$
$\displaystyle S_n = n a_1 \;\text{if}\; r = 1$
1.1.3
Writing Geometric Series in Sigma Notation
Write the general term $a_1 r^{k-1}$ and let $k$ run from 1 to $n$ . Find $n$ by solving $a_1 r^{n-1} =$ last term.
e.g. $3 + 6 + 12 + \dots + 96 = \sum_{k=1}^{6} 3 \cdot 2^{k-1}$ , since $3 \cdot 2^5 = 96$
$\displaystyle \sum_{k=1}^{n} a_1 r^{k-1}$
1.1.4
Sums of Finite Geometric Series Given in Sigma Notation
Plug the starting index into the term to get the first term $a$ . The ratio $r$ is the base of the exponent. The number of terms is top index minus bottom index plus 1.
e.g. $\sum_{k=0}^{4} 2^k$ : $a = 1$ , $r = 2$ , $n = 5$ , so $S = \frac{1 - 32}{1 - 2} = 31$
$\displaystyle n = \text{top} - \text{bottom} + 1$
$\displaystyle S = \frac{a(1 - r^n)}{1 - r}$
1.1.5
Solving Geometric Series Problems Using Exponential Equations and Inequalities
To find how many terms reach a target, set the sum formula equal to (or greater than) the target, isolate $r^n$ , and take logs. For "at least" questions, round $n$ up. Dividing by the log of a number less than 1 (a negative number) flips the inequality.
e.g. $1 + 2 + 4 + \dots > 1000$ : $2^n - 1 > 1000 \Rightarrow n > \log_2 1001 \approx 9.97$ , so $n = 10$
$\displaystyle \frac{a(1 - r^n)}{1 - r} \ge T$
$\displaystyle \Rightarrow\; \text{isolate } r^n \text{, take logs}$
1.1.6
Modeling With Geometric Series
Use a geometric series when an amount changes by the same percent each step and you need the running total. The first amount is $a$ ; the ratio is 1 plus the growth rate, or 1 minus the decay rate.
e.g. A ball falls 10 ft, then each fall is half the last. First 4 falls: $\frac{10(1 - 0.5^4)}{1 - 0.5} = 18.75$ ft
$\displaystyle r = 1 \pm \text{rate}$
$\displaystyle S_n = \frac{a(1 - r^n)}{1 - r}$
1.1.7
Modeling Financial Problems Using Geometric Series
Equal deposits $P$ made at the end of each period, earning rate $i$ per period, form a geometric series with ratio $1 + i$ . Its sum is the future value after $n$ deposits.
e.g. $100 at the end of each year for 3 years at 10% per year: $100 + 110 + 121 = 100 \cdot \frac{1.1^3 - 1}{0.1} = \$331$
$\displaystyle FV = P \cdot \frac{(1 + i)^n - 1}{i}$

Unit 2 19 topics

Polynomials

Dividing Polynomials, Polynomial Theorems, and Graphs of Polynomials.

2.2 Dividing Polynomials

2.2.1
Dividing Polynomials Using Synthetic Division
To divide by $x - c$ , write $c$ and the coefficients (0 for any missing power). Bring down the first, multiply by $c$ , add to the next column, and repeat. The last number is the remainder; the rest are the quotient's coefficients, one degree lower.
e.g. $(x^3 - 2x^2 - 5x + 6) \div (x - 3)$ : $1, -2, -5, 6$ becomes $1, 1, -2, 0$ , so $x^2 + x - 2$ , remainder 0
$\displaystyle p(x) = (x - c)\,q(x) + R$
2.2.2
Dividing Polynomials by Linear Binomials Using Long Division
Divide the leading terms, multiply the divisor by the result, subtract, and bring down the next term. Repeat until only a constant remains. Write 0 terms for missing powers.
e.g. $(2x^2 + 3x - 4) \div (x + 2) = 2x - 1 - \tfrac{2}{x + 2}$
$\displaystyle \frac{p(x)}{d(x)} = q(x) + \frac{R}{d(x)}$
2.2.3
Dividing Polynomials Using Long Division
The same steps work for any divisor, such as a quadratic. Stop when the remainder's degree is less than the divisor's degree, so the remainder can be linear.
e.g. $(x^3 + 2x^2 + 3x + 4) \div (x^2 + 1) = x + 2$ , remainder $2x + 2$
$\displaystyle p(x) = d(x)\,q(x) + r(x)$
$\displaystyle \deg r < \deg d$
2.2.4
Dividing Polynomials by Manipulating Rational Expressions
Rewrite the numerator so it contains the denominator, by adding and subtracting the same amount. Then split the fraction.
e.g. $\frac{x^2}{x - 1} = \frac{(x^2 - 1) + 1}{x - 1} = x + 1 + \tfrac{1}{x - 1}$
$\displaystyle \frac{x + 5}{x + 2} = \frac{(x + 2) + 3}{x + 2}$
$\displaystyle = 1 + \frac{3}{x + 2}$
2.2.5
Closure Properties of Polynomials
Polynomials are closed under addition, subtraction, and multiplication: the result is always a polynomial. They are not closed under division, since a quotient like $\frac{1}{x}$ is not a polynomial.
e.g. $(x + 1)(x - 2) = x^2 - x - 2$ is a polynomial; $\frac{x}{x + 1}$ is not

2.3 Polynomial Theorems

2.3.1
The Factor Theorem
$(x - c)$ is a factor of $p(x)$ exactly when $p(c) = 0$ .
e.g. $p(x) = x^3 - 8$ : $p(2) = 0$ , so $x - 2$ is a factor
$\displaystyle p(c) = 0 \iff (x - c) \text{ is a factor}$
2.3.2
Determining Polynomial Coefficients Using the Factor Theorem
If $(x - c)$ is a factor, substitute $x = c$ , set the result equal to 0, and solve for the unknown coefficient. For a factor $(ax - b)$ , use $x = \frac{b}{a}$ .
e.g. $x^3 + kx - 6$ has factor $x - 2$ : $8 + 2k - 6 = 0 \Rightarrow k = -1$
$\displaystyle p(c) = 0$
2.3.3
Factoring Cubic Polynomials Using the Factor Theorem
Test small divisors of the constant term until $p(c) = 0$ . Divide by $(x - c)$ , then factor the remaining quadratic.
e.g. $x^3 - 6x^2 + 11x - 6$ : $p(1) = 0$ , so $(x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)$
$\displaystyle p(x) = (x - c)\,q(x)$
2.3.4
Factoring Quartic Polynomials Using the Factor Theorem
Find one root $c_1$ and divide to get a cubic. Find a root $c_2$ of the cubic and divide again, then factor the quadratic left over.
e.g. $x^4 - 5x^3 + 5x^2 + 5x - 6$ : $p(1) = p(-1) = 0$ , so it equals $(x - 1)(x + 1)(x - 2)(x - 3)$
$\displaystyle p(x) = (x - c_1)(x - c_2)\,q(x)$
2.3.5
Multiplicities of the Roots of Polynomials
A root's multiplicity is the number of times its factor appears. At a root of odd multiplicity the graph crosses the $x$ -axis. At even multiplicity it touches and turns back.
e.g. $(x - 1)^2(x + 3)$ : $x = 1$ has multiplicity 2 (touches), $x = -3$ has multiplicity 1 (crosses)
$\displaystyle p(x) = (x - c)^m\,q(x)$
$\displaystyle q(c) \ne 0$
2.3.6
Finding Multiplicities of the Roots of Quartic Polynomials by Factoring
Factor completely, group repeated factors as powers, and read each exponent. For a quartic the multiplicities of the real roots add up to at most 4.
e.g. $x^4 - 2x^3 + x^2 = x^2(x - 1)^2$ : roots 0 and 1, each with multiplicity 2
2.3.7
The Remainder Theorem
When $p(x)$ is divided by $(x - c)$ , the remainder is $p(c)$ .
e.g. $(x^3 - 2x + 5) \div (x - 2)$ : remainder $p(2) = 8 - 4 + 5 = 9$
$\displaystyle p(x) = (x - c)\,q(x) + p(c)$
2.3.8
The Rational Roots Theorem
If a polynomial has integer coefficients, every rational root $\frac{p}{q}$ in lowest terms has $p$ dividing the constant term $a_0$ and $q$ dividing the leading coefficient $a_n$ . This gives a list of candidates to test.
e.g. $2x^3 - 3x^2 - 3x + 2$ : candidates $\pm1, \pm2, \pm\tfrac12$ ; $p(2) = 16 - 12 - 6 + 2 = 0$ , so 2 is a root
$\displaystyle \pm\frac{\text{factors of } a_0}{\text{factors of } a_n}$

2.4 Graphs of Polynomials

A root of odd multiplicity crosses the x-axis and a root of even multiplicity only touches it. The degree and the sign of the leading coefficient fix which way the two ends point.

2.4.1
Graphing Elementary Cubic Functions
$y = x^3$ passes through the origin, rises from left to right, and is symmetric about the origin. In the transformed form the center point moves to $(h, k)$ and a negative $a$ flips the graph.
e.g. $y = (x - 1)^3 + 2$ : center $(1, 2)$ , also through $(2, 3)$ and $(0, 1)$
$\displaystyle y = a(x - h)^3 + k$
2.4.2
Graphing Cubic Curves Containing Three Distinct Real Roots
The graph crosses the $x$ -axis at each root. If $a > 0$ it falls on the left and rises on the right; if $a < 0$ the reverse. Plug in $x = 0$ for the $y$ -intercept.
e.g. $y = (x + 1)(x - 1)(x - 2)$ : crosses at $-1, 1, 2$ ; $y$ -intercept 2
$\displaystyle y = a(x - r_1)(x - r_2)(x - r_3)$
2.4.3
Graphing Cubic Curves Containing a Double Root
The squared factor gives a root where the graph touches the $x$ -axis and turns back. The graph crosses at the single root.
e.g. $y = (x - 1)^2(x + 2)$ : touches at $x = 1$ , crosses at $x = -2$ , $y$ -intercept 2
$\displaystyle y = a(x - r)^2(x - s)$
2.4.4
Graphing Cubic Curves Containing One Distinct Real Root
The cubic factors as a linear factor times a quadratic with no real roots, so the graph crosses the $x$ -axis only once.
e.g. $y = (x - 2)(x^2 + 1)$ : crosses only at $x = 2$ , $y$ -intercept $-2$
$\displaystyle y = a(x - r)(x^2 + bx + c)$
$\displaystyle b^2 - 4c < 0$
2.4.5
End Behavior of Polynomials
For large $\lvert x \rvert$ the leading term $a_n x^n$ controls the graph. Even degree: both ends point the same way, up if $a_n > 0$ . Odd degree: the ends point opposite ways, rising on the right if $a_n > 0$ .
e.g. $f(x) = -2x^3 + x$ : as $x \to \infty$ , $f(x) \to -\infty$ ; as $x \to -\infty$ , $f(x) \to \infty$
2.4.6
Graphing General Polynomials
Factor to find the roots and their multiplicities, find the $y$ -intercept, and use the end behavior. Cross at odd multiplicity, touch at even, and connect smoothly. A degree $n$ polynomial has at most $n - 1$ turning points.
e.g. $y = x(x - 2)^2(x + 1)$ : degree 4 with $a > 0$ , so both ends up; crosses at $-1$ and 0, touches at 2

Unit 3 21 topics

Rational Equations & Functions

Rational Equations, Reciprocal Functions, and Rational Functions.

3.5 Rational Equations

Cross-multiplying is multiplying both sides by both denominators at once. It only works when each side is a single fraction.

3.5.1
Rational Equations With Three Terms
Multiply every term by the LCD to clear the fractions, then solve. Reject any answer that makes a denominator zero.
e.g. $\frac{3}{x} - \frac{1}{2} = \frac{1}{x}$ : multiply by $2x$ : $6 - x = 2$ , so $x = 4$
3.5.2
Advanced Rational Equations
Factor every denominator first to find the LCD and the excluded values. Multiply through, solve the resulting equation (often a quadratic), and discard extraneous roots.
e.g. $\frac{x}{x - 2} - \frac{1}{x + 2} = \frac{8}{x^2 - 4}$ gives $x^2 + x - 6 = 0$ , so $x = -3$ ( $x = 2$ is extraneous)
3.5.3
Further Advanced Rational Equations
Use the same method with more terms or harder denominators. List the excluded values before solving and check every answer against them. If every answer is excluded, there is no solution.
e.g. $\frac{2}{x - 1} + \frac{1}{x + 1} = \frac{4}{x^2 - 1}$ : $2(x + 1) + (x - 1) = 4 \Rightarrow x = 1$ , which is excluded, so no solution

3.6 Reciprocal Functions

3.6.1
Graphing Reciprocal Functions
$y = \frac{k}{x}$ is a hyperbola with asymptotes $x = 0$ and $y = 0$ . If $k > 0$ the branches are in quadrants I and III; if $k < 0$ they are in II and IV.
e.g. $y = \frac{4}{x}$ passes through $(1, 4)$ , $(2, 2)$ , and $(-2, -2)$
$\displaystyle y = \frac{k}{x}$
3.6.2
Graph Transformations of Reciprocal Functions
$h$ shifts the graph right, $k$ shifts it up, $\lvert a \rvert$ stretches it, and a negative $a$ reflects it. The asymptotes move with the shifts.
e.g. $y = \frac{1}{x + 3}$ is $\frac{1}{x}$ shifted 3 left, with vertical asymptote $x = -3$
$\displaystyle y = \frac{a}{x - h} + k$
$\displaystyle \text{asymptotes: } x = h,\; y = k$
3.6.3
Combining Graph Transformations of Reciprocal Functions
Apply the stretch and reflection first, then the shifts. If the function is given as $\frac{px + q}{x + r}$ , divide first to get the form $\frac{a}{x - h} + k$ .
e.g. $y = -\frac{2}{x - 1} + 4$ : stretch by 2, reflect, shift 1 right and 4 up; center $(1, 4)$
$\displaystyle \frac{2x + 7}{x + 3} = 2 + \frac{1}{x + 3}$
3.6.4
Domain and Range of Transformed Reciprocal Functions
For $y = \frac{a}{x - h} + k$ , the domain is all real numbers except $h$ and the range is all real numbers except $k$ .
e.g. $y = \frac{3}{x + 2} - 5$ : domain $x \ne -2$ , range $y \ne -5$
$\displaystyle \text{Domain: } x \ne h$
$\displaystyle \text{Range: } y \ne k$
3.6.5
Inverses of Reciprocal Functions
Swap $x$ and $y$ , then solve for $y$ . The inverse of $y = \frac{a}{x - h} + k$ has the asymptotes swapped. $y = \frac{k}{x}$ is its own inverse.
e.g. $f(x) = \frac{2}{x} + 1 \Rightarrow f^{-1}(x) = \frac{2}{x - 1}$
$\displaystyle f^{-1}(x) = \frac{a}{x - k} + h$
3.6.6
Finding Intersections of Lines and Reciprocal Functions
Set the expressions equal, multiply by the denominator to get a quadratic, and solve. Two real solutions means two intersections, one means the line is tangent, none means they do not meet.
e.g. $\frac{6}{x} = x + 1 \Rightarrow x^2 + x - 6 = 0$ , so the points are $(2, 3)$ and $(-3, -2)$

3.7 Rational Functions

A factor that cancels leaves a hole; a zero of the denominator that remains becomes a vertical asymptote. With equal degrees, the horizontal asymptote is the ratio of leading coefficients.

3.7.1
Roots of Rational Functions
The zeros are the $x$ -values that make the numerator 0 but not the denominator. Cancel common factors first: a cancelled factor gives a hole, not a root.
e.g. $f(x) = \frac{x^2 - 4}{x - 2}$ : root only at $x = -2$ (there is a hole at $x = 2$ )
$\displaystyle \frac{p(x)}{q(x)} = 0$
$\displaystyle \Rightarrow\; p(x) = 0 \text{ and } q(x) \ne 0$
3.7.2
Vertical Asymptotes of Rational Functions
Cancel common factors first. Each $x$ -value that still makes the denominator zero is a vertical asymptote.
e.g. $f(x) = \frac{x + 1}{(x - 3)(x + 1)}$ : vertical asymptote $x = 3$ only
3.7.3
Locating Holes in Rational Functions
A factor that cancels completely from the numerator and denominator gives a hole. Its $x$ -value makes that factor zero; its $y$ -value comes from the simplified function. If the factor still remains in the denominator, it is an asymptote instead.
e.g. $f(x) = \frac{x^2 - 1}{x - 1} = x + 1$ for $x \ne 1$ : hole at $(1, 2)$
3.7.4
Horizontal Asymptotes of Rational Functions
Compare the degree $n$ of the numerator with the degree $m$ of the denominator. Leading coefficients are $a_n$ and $b_m$ .
e.g. $\frac{3x^2 + 1}{2x^2 - x}$ : horizontal asymptote $y = \tfrac32$
$\displaystyle n < m: \; y = 0$
$\displaystyle n = m: \; y = \frac{a_n}{b_m}$
$\displaystyle n > m: \; \text{none}$
3.7.5
End Behavior of Rational Functions
Divide to write $f$ as a polynomial $q(x)$ plus a proper fraction. The fraction goes to 0 as $x \to \pm\infty$ , so $f$ behaves like $q(x)$ . If the numerator's degree is one more than the denominator's, $q$ is linear and gives a slant asymptote.
e.g. $\frac{x^2 + 1}{x} = x + \frac{1}{x}$ : slant asymptote $y = x$
$\displaystyle f(x) = q(x) + \frac{r(x)}{d(x)}$
$\displaystyle \Rightarrow\; f(x) \approx q(x) \text{ for large } \lvert x \rvert$
3.7.6
Infinite Limits of Rational Functions
Near a vertical asymptote $x = a$ , the function goes to $\infty$ or $-\infty$ . Find the sign by testing a value just left and just right of $a$ .
e.g. $f(x) = \frac{1}{x - 2}$ : $\lim_{x \to 2^+} f(x) = \infty$ , $\lim_{x \to 2^-} f(x) = -\infty$
$\displaystyle \lim_{x \to a^-} f(x) = \pm\infty$
$\displaystyle \lim_{x \to a^+} f(x) = \pm\infty$
3.7.7
Infinite Limits of Rational Functions: Advanced Cases
Find the sign of each factor near $a$ . If $(x - a)$ has an even power in the denominator, both one-sided limits have the same sign. An odd power gives opposite signs.
e.g. $f(x) = \frac{x - 3}{(x - 1)^2}$ near $x = 1$ : numerator near $-2$ , denominator small and positive, so $f \to -\infty$ from both sides
3.7.8
The Domain and Range of a Rational Function
The domain is all real numbers except the zeros of the original denominator, including holes. For the range, solve $y = f(x)$ for $x$ and exclude the $y$ -values with no solution. Also exclude a hole's $y$ -value if the graph does not reach it elsewhere.
e.g. $f(x) = \frac{2x + 1}{x - 3}$ : $x = \frac{3y + 1}{y - 2}$ , so domain $x \ne 3$ , range $y \ne 2$
3.7.9
Identifying a Rational Function From a Graph
Each vertical asymptote $x = a$ gives a factor $(x - a)$ in the denominator. Each $x$ -intercept $r$ gives a factor $(x - r)$ in the numerator. Use the horizontal asymptote or a known point to find the constant $k$ .
e.g. Vertical asymptote $x = 1$ , $x$ -intercept $-2$ , horizontal asymptote $y = 3$ : $f(x) = \frac{3(x + 2)}{x - 1}$
$\displaystyle f(x) = \frac{k(x - r)}{x - a}$
3.7.10
Identifying a Rational Function From a Graph Containing Holes
A hole at $x = c$ means $(x - c)$ is a factor of both the numerator and the denominator. Build the function from the asymptotes and intercepts, then multiply top and bottom by $(x - c)$ .
e.g. The graph of $y = \frac{2}{x}$ with a hole at $(1, 2)$ : $f(x) = \frac{2(x - 1)}{x(x - 1)}$
$\displaystyle f(x) = \frac{g(x)\,(x - c)}{h(x)\,(x - c)}$
3.7.11
Identifying the Graph of a Rational Function
Find the features from the equation and match them to a graph: zeros, vertical asymptotes, holes, the horizontal or slant asymptote, the $y$ -intercept, and the sign between key $x$ -values.
e.g. $f(x) = \frac{x - 1}{x + 2}$ : zero at $x = 1$ , asymptotes $x = -2$ and $y = 1$ , $y$ -intercept $-\tfrac12$
3.7.12
Closure Properties of Rational Expressions
Rational expressions are closed under addition, subtraction, multiplication, and division by a nonzero rational expression: the result is always a ratio of polynomials.
e.g. $\frac{1}{x} + \frac{1}{x + 1} = \frac{2x + 1}{x(x + 1)}$ , still rational
$\displaystyle \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$
$\displaystyle \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$

Unit 4 13 topics

Radical Functions

Radical Expressions, and Graphs of Radical Functions.

4.8 Radical Expressions

A root is a fractional power. Even roots need an absolute value when the sign of x is unknown; odd roots do not.

4.8.1
Simplifying Square Root Expressions Using Polynomial Factorization
Factor the radicand to find a perfect-square factor, then take its square root out. The root of a squared expression is its absolute value unless that expression is known to be nonnegative.
e.g. $\sqrt{4x^2 - 8x + 4} = \sqrt{4(x-1)^2} = 2\lvert x - 1 \rvert$
$\displaystyle \sqrt{A^2 B} = \lvert A \rvert \sqrt{B}$
$\displaystyle \sqrt{x^2 + 6x + 9} = \lvert x + 3 \rvert$
4.8.2
Rationalizing Denominators of Algebraic Expressions
Multiply top and bottom by the root in the denominator.
$\displaystyle \frac{a}{\sqrt{x}} = \frac{a\sqrt{x}}{x}$
4.8.3
Rationalizing Denominators With Two Terms
Multiply by the conjugate: the same two terms with the middle sign flipped. The difference of squares clears the root.
e.g. $\tfrac{1}{3 + \sqrt2} = \tfrac{3 - \sqrt2}{7}$
$\displaystyle \frac{1}{a + \sqrt{b}} \cdot \frac{a - \sqrt{b}}{a - \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b}$
4.8.4
Further Rationalizing Denominators of Algebraic Expressions
For a cube root, multiply by enough to make a perfect cube.
$\displaystyle \frac{1}{\sqrt[3]{x}} = \frac{\sqrt[3]{x^2}}{x}$

4.9 Graphs of Radical Functions

4.9.1
Graphing the Square Root Function
The graph of $y = \sqrt{x}$ starts at the origin and rises to the right, flattening as it goes. Domain $x \ge 0$ , range $y \ge 0$ .
e.g. Plot $(0,0)$ , $(1,1)$ , $(4,2)$ , $(9,3)$ and connect with a smooth curve.
$\displaystyle y = \sqrt{x}$
4.9.2
Graph Transformations of Square Root Functions
The graph of $\sqrt{x}$ shifts right $h$ and up $k$ , so the starting point moves to $(h, k)$ . $\lvert a \rvert$ stretches vertically and $a < 0$ reflects over the $x$ -axis. $b < 0$ flips the graph so it opens to the left.
e.g. $y = \sqrt{x - 2} + 1$ starts at $(2, 1)$ and passes through $(3, 2)$ and $(6, 3)$ .
$\displaystyle y = a\sqrt{b(x - h)} + k$
4.9.3
Graphing the Cube Root Function
The graph of $y = \sqrt[3]{x}$ passes through the origin and extends in both directions. Domain and range are all real numbers. It is symmetric about the origin.
e.g. Plot $(-8,-2)$ , $(-1,-1)$ , $(0,0)$ , $(1,1)$ , $(8,2)$ .
$\displaystyle y = \sqrt[3]{x}$
4.9.4
Properties of Transformed Square Root Functions
For $y = a\sqrt{x - h} + k$ the endpoint is $(h, k)$ and the domain is $x \ge h$ . If $a > 0$ the graph increases and the range is $y \ge k$ . If $a < 0$ it decreases and the range is $y \le k$ .
e.g. $y = -2\sqrt{x + 1} + 3$ : endpoint $(-1, 3)$ , domain $x \ge -1$ , range $y \le 3$ , decreasing
$\displaystyle y = a\sqrt{x - h} + k$
4.9.5
The Domain of a Transformed Radical Function
For an even root, the radicand must be at least 0, so solve radicand $\ge 0$ . For an odd root, the domain is all real numbers.
e.g. $y = \sqrt{6 - 2x}$ : $6 - 2x \ge 0 \Rightarrow x \le 3$
$\displaystyle \sqrt{u}:\; u \ge 0$
$\displaystyle \sqrt[3]{u}:\; \text{all reals}$
4.9.6
The Range of a Transformed Radical Function
For $y = a\sqrt{b(x - h)} + k$ , the range starts at $k$ . If $a > 0$ the range is $y \ge k$ ; if $a < 0$ it is $y \le k$ . A transformed cube root has all real numbers as its range.
e.g. $y = -\sqrt{x - 4} + 5$ : range $y \le 5$
$\displaystyle a > 0:\; y \ge k$
$\displaystyle a < 0:\; y \le k$
4.9.7
Roots of Transformed Radical Functions
Set $y = 0$ , isolate the radical, then square (or cube) both sides. A square root can never equal a negative number, so check every answer in the original equation.
e.g. $\sqrt{x - 1} - 2 = 0 \Rightarrow \sqrt{x - 1} = 2 \Rightarrow x = 5$
$\displaystyle a\sqrt{x - h} + k = 0$
$\displaystyle \Rightarrow\; \sqrt{x - h} = -\tfrac{k}{a}$
4.9.8
Inverses of Radical Functions
Swap $x$ and $y$ , then solve for $y$ . The inverse of a square root function is half a parabola: its domain is the range of the original. A cube root inverse needs no restriction.
e.g. $f(x) = \sqrt[3]{x} + 1 \Rightarrow f^{-1}(x) = (x - 1)^3$
$\displaystyle f(x) = \sqrt{x - 2},\; x \ge 2$
$\displaystyle \Rightarrow\; f^{-1}(x) = x^2 + 2,\; x \ge 0$
4.9.9
Finding Intersections of Lines and Radical Functions
Set the two expressions equal, isolate the radical, square both sides, and solve. Squaring can create extraneous solutions, so check each one in the original equation.
e.g. $x = -1$ fails ( $\sqrt1 \ne -1$ ), so the only intersection is $(2, 2)$ .
$\displaystyle \sqrt{x + 2} = x$
$\displaystyle \Rightarrow\; x + 2 = x^2$
$\displaystyle \Rightarrow\; (x - 2)(x + 1) = 0$

Unit 5 21 topics

Inequalities

Quadratic Inequalities, Polynomial Inequalities, and Non-Polynomial Inequalities.

5.10 Quadratic Inequalities

The product is positive where the parabola is above the x-axis. The roots split the line into regions; test one point in each, and the answer is x < 1 or x > 4.

5.10.1
Solving Elementary Quadratic Inequalities
For $c > 0$ : $x^2 < c$ gives the values between the roots, and $x^2 > c$ gives the values outside them. If $c < 0$ , $x^2 < c$ has no solution and $x^2 > c$ holds for all $x$ .
e.g. $x^2 \le 9 \Rightarrow -3 \le x \le 3$
$\displaystyle x^2 < c \;\Rightarrow\; -\sqrt{c} < x < \sqrt{c}$
$\displaystyle x^2 > c \;\Rightarrow\; x < -\sqrt{c}$
$\displaystyle \text{or}\; x > \sqrt{c}$
5.10.2
Solving Quadratic Inequalities From Graphs
$f(x) > 0$ where the graph is above the $x$ -axis, and $f(x) < 0$ where it is below. The $x$ -intercepts are the boundaries. Include them only for $\le$ or $\ge$ .
e.g. $y = (x - 1)(x - 5)$ opens up, so $y < 0$ for $1 < x < 5$ .
5.10.3
Solving Quadratic Inequalities Using the Graphical Method
Move everything to one side, find the roots, and sketch the parabola. If it opens up, it is negative between the roots and positive outside them. If it opens down, the reverse.
e.g. $x^2 - 2x - 8 > 0 \Rightarrow (x - 4)(x + 2) > 0 \Rightarrow x < -2$ or $x > 4$
$\displaystyle a > 0:\; ax^2 + bx + c < 0$
$\displaystyle \text{between the roots}$
5.10.4
Solving Quadratic Inequalities Using the Sign Table Method
Move everything to one side and factor. Mark the roots on a number line, find the sign of each factor on each interval, and multiply the signs. Keep the intervals with the sign you need.
e.g. $-x^2 + x + 6 \ge 0 \Rightarrow (x - 3)(x + 2) \le 0 \Rightarrow -2 \le x \le 3$
5.10.5
Determining Unknown Parameters in Quadratic Equations With Real Solutions
$ax^2 + bx + c = 0$ has real solutions when the discriminant is at least 0, and two distinct real solutions when it is positive. Solve that inequality for the parameter.
e.g. $x^2 + kx + 9 = 0$ : $k^2 - 36 \ge 0 \Rightarrow k \le -6$ or $k \ge 6$
$\displaystyle b^2 - 4ac \ge 0$
5.10.6
Determining Unknown Parameters in Quadratic Equations With No Real Solutions
A quadratic has no real solutions when its discriminant is negative. Solve that inequality for the parameter.
e.g. $x^2 + 4x + k = 0$ : $16 - 4k < 0 \Rightarrow k > 4$
$\displaystyle b^2 - 4ac < 0$
5.10.7
Determining Unknown Parameters in Quadratic Equations: Advanced Cases
If the parameter is in the $x^2$ coefficient, the equation is only quadratic when that coefficient is not 0, so check that case separately. A quadratic is positive for all $x$ when it opens up and has no real roots.
e.g. $kx^2 + 4x + 1 = 0$ has two real solutions when $k \ne 0$ and $16 - 4k > 0$ , so $k < 4$ , $k \ne 0$ .
$\displaystyle a \ne 0 \;\text{and}\; b^2 - 4ac > 0$
$\displaystyle ax^2 + bx + c > 0 \;\text{for all } x$
$\displaystyle \iff a > 0,\; b^2 - 4ac < 0$

5.11 Polynomial Inequalities

5.11.1
Inequalities Involving Powers of Binomials
An even power $(x - a)^n$ is never negative and equals 0 only at $x = a$ . An odd power $(x - a)^n$ has the same sign as $x - a$ .
e.g. $(x - 3)^4 > 0 \Rightarrow x \ne 3$ ; $(x - 2)^5 < 0 \Rightarrow x < 2$
$\displaystyle (x - a)^{\text{even}} \ge 0$
$\displaystyle (x - a)^{\text{odd}} \text{ has the sign of } x - a$
5.11.2
Solving Polynomial Inequalities Using a Graphical Method
Move everything to one side and factor. Plot the zeros, use the end behavior, and remember the graph crosses at odd-multiplicity zeros and bounces at even ones. Read off where the graph is above or below the axis.
e.g. $x(x - 2)(x + 3) > 0$ : zeros $-3, 0, 2$ , so $-3 < x < 0$ or $x > 2$
5.11.3
Solving Polynomial Inequalities Using Special Factoring Techniques and the Graphical
Factor with grouping, difference of squares, or sum and difference of cubes, then use the graph's zeros and end behavior to read off the solution.
e.g. $x^3 - x^2 - 4x + 4 \le 0 \Rightarrow (x - 1)(x - 2)(x + 2) \le 0 \Rightarrow x \le -2$ or $1 \le x \le 2$
$\displaystyle a^2 - b^2 = (a - b)(a + b)$
$\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
$\displaystyle a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
5.11.4
Solving Polynomial Inequalities Using the Sign Table Method
Factor completely and list the zeros. Make a sign table with one row per factor and multiply the signs on each interval. A factor with an even power never changes sign.
e.g. $(x - 1)^2(x + 4) < 0$ : $(x-1)^2 > 0$ for $x \ne 1$ , so $x + 4 < 0 \Rightarrow x < -4$

5.12 Non-Polynomial Inequalities

5.12.1
Solving Radical Inequalities
First require the radicand to be at least 0. When both sides are nonnegative, square both sides. Combine the conditions.
e.g. $\sqrt{x - 3} < 2 \Rightarrow 0 \le x - 3 < 4 \Rightarrow 3 \le x < 7$
$\displaystyle \sqrt{A} < B,\; B > 0$
$\displaystyle \Rightarrow\; 0 \le A < B^2$
5.12.2
Solving Inequalities Involving Exponential Functions
Write both sides with the same base, then compare exponents. If the base is greater than 1 keep the direction; if it is between 0 and 1, flip it. You can also take logs of both sides.
e.g. $\left(\tfrac13\right)^x \ge 9 \Rightarrow 3^{-x} \ge 3^2 \Rightarrow x \le -2$
$\displaystyle b > 1:\; b^u > b^v \iff u > v$
$\displaystyle 0 < b < 1:\; b^u > b^v \iff u < v$
5.12.3
Solving Inequalities Involving Logarithmic Functions
The argument of a log must be positive, so include that condition. Rewrite in exponential form. Keep the direction for a base greater than 1; flip it for a base between 0 and 1.
e.g. $\log_2(x - 1) < 3 \Rightarrow 0 < x - 1 < 8 \Rightarrow 1 < x < 9$
$\displaystyle b > 1:\; \log_b u < c$
$\displaystyle \iff 0 < u < b^c$
5.12.4
Solving Inequalities Involving Exponential Functions and Polynomials
An exponential like $e^x$ or $2^x$ is always positive, so you can divide by it without flipping the sign. For expressions with $4^x$ and $2^x$ , substitute $u = 2^x$ , solve in $u$ , then convert back.
e.g. $e^x(x^2 - 4) < 0 \Rightarrow x^2 - 4 < 0 \Rightarrow -2 < x < 2$
$\displaystyle b^x > 0 \;\text{for all } x$
5.12.5
Solving Inequalities Involving Positive and Negative Factors
A product or quotient is positive when its factors have the same sign and negative when they have opposite signs. A factor that is always positive, like $x^2 + 1$ or $e^x$ , does not affect the sign.
e.g. $(x^2 + 1)(x - 5) \le 0 \Rightarrow x - 5 \le 0 \Rightarrow x \le 5$
$\displaystyle AB > 0 \iff A, B \text{ same sign}$
$\displaystyle AB < 0 \iff A, B \text{ opposite signs}$
5.12.6
Solving Inequalities Involving Geometric Sequences
Write the term or sum formula, set up the inequality, and solve for $n$ using logs or by testing powers. $n$ must be a positive integer, so take the first one that works. Dividing by the log of a number less than 1 flips the sign.
e.g. First term of $3, 6, 12, \ldots$ over 1000: $3 \cdot 2^{n-1} > 1000 \Rightarrow 2^{n-1} > 333.\overline{3} \Rightarrow n = 10$
$\displaystyle a_n = a_1 r^{n-1}$
$\displaystyle S_n = \frac{a_1(1 - r^n)}{1 - r}$
5.12.7
Solving Rational Inequalities
Get 0 on one side and a single fraction on the other. The zeros of the numerator and denominator split the number line. Test each interval. Never include a zero of the denominator, and do not multiply by a denominator whose sign is unknown.
e.g. $\dfrac{x - 2}{x + 3} \ge 0 \Rightarrow x < -3$ or $x \ge 2$
5.12.8
Further Solving of Rational Inequalities
Subtract to get 0 on one side, combine over a common denominator, and simplify the numerator. Then use a sign table on the critical values.
e.g. Critical values 1 and 3, so $x < 1$ or $x > 3$ .
$\displaystyle \frac{2}{x - 1} < 1$
$\displaystyle \Rightarrow\; \frac{3 - x}{x - 1} < 0$
5.12.9
Solving Two-Variable Nonlinear Inequalities
Graph the boundary curve: dashed for $<$ or $>$ , solid for $\le$ or $\ge$ . Test a point not on the curve and shade the side where it makes the inequality true.
e.g. $y > x^2 - 4$ : dashed parabola; $(0,0)$ gives $0 > -4$ , true, so shade above.
5.12.10
Further Solving of Two-Variable Nonlinear Inequalities
For a system, graph each inequality and shade the region where all of them hold. A point is a solution only if it satisfies every inequality.
e.g. Is $(1, 3)$ in $y \ge x^2$ , $x^2 + y^2 < 16$ ? $3 \ge 1$ and $1 + 9 < 16$ , so yes.

Unit 6 20 topics

Conics

Ellipses as Conic Sections, and Hyperbolas as Conic Sections.

6.13 Ellipses as Conic Sections

For an ellipse, a is half the long axis, b is half the short axis, and the foci sit c from the center with c² = a² − b². From any point on the curve, the two distances to the foci add to 2a.

6.13.1
Introduction to Ellipses
An ellipse is the set of points whose distances to two fixed points, the foci, add to a constant $2a$ . The major axis is the longest diameter and the minor axis is the shortest. The center is their midpoint.
e.g. Foci $(\pm 3, 0)$ and distance sum 10 give $a = 5$ .
$\displaystyle d_1 + d_2 = 2a$
6.13.2
Equations of Ellipses Centered at the Origin
Here $a > b > 0$ , where $a$ is half the major axis and $b$ is half the minor axis. The larger denominator tells you the direction of the major axis.
e.g. $\tfrac{x^2}{25} + \tfrac{y^2}{9} = 1$ : horizontal, vertices $(\pm 5, 0)$ , co-vertices $(0, \pm 3)$
$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \;\text{horizontal}$
$\displaystyle \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \;\text{vertical}$
6.13.3
Equations of Ellipses Centered at a General Point
Replace $x$ with $x - h$ and $y$ with $y - k$ to move the center to $(h, k)$ . The axes keep their lengths $2a$ and $2b$ .
e.g. $\tfrac{(x-2)^2}{16} + \tfrac{(y+1)^2}{4} = 1$ : center $(2, -1)$ , vertices $(-2, -1)$ and $(6, -1)$
$\displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$
6.13.4
Finding the Center and Axes of Ellipses by Completing the Square
Group the $x$ terms and the $y$ terms, factor out each squared term's coefficient, and complete the square in each group. Add the same amounts to the other side, then divide so the right side is 1.
e.g. $4x^2 + 9y^2 - 16x + 18y = 11 \Rightarrow 4(x-2)^2 + 9(y+1)^2 = 36$ : center $(2, -1)$ , $a = 3$ , $b = 2$
$\displaystyle Ax^2 + Cy^2 + Dx + Ey + F = 0$
$\displaystyle \Rightarrow\; \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$
6.13.5
Finding Intercepts of Ellipses
For $x$ -intercepts set $y = 0$ ; for $y$ -intercepts set $x = 0$ . Solve each, and note that an ellipse may miss an axis entirely. Centered at the origin, the intercepts are the vertices and co-vertices.
e.g. $\tfrac{(x-1)^2}{4} + \tfrac{y^2}{9} = 1$ : $y = 0 \Rightarrow (x - 1)^2 = 4 \Rightarrow x = -1, 3$
$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
$\displaystyle \Rightarrow\; (\pm a, 0),\; (0, \pm b)$
6.13.6
Finding Intersections of Ellipses and Lines
Substitute the line into the ellipse and solve the quadratic. Two real solutions mean two intersection points, one means the line is tangent, and none means they do not meet.
e.g. $(5x - 6)(x - 2) = 0$ , so the points are $(2, 0)$ and $\left(\tfrac65, -\tfrac45\right)$ .
$\displaystyle x^2 + 4y^2 = 4,\; y = x - 2$
$\displaystyle \Rightarrow\; 5x^2 - 16x + 12 = 0$
6.13.7
Foci of Ellipses
The foci lie on the major axis, $c$ units from the center on each side, where $a$ is the semi-major axis.
e.g. $\tfrac{x^2}{25} + \tfrac{y^2}{16} = 1$ : $c^2 = 9$ , foci $(\pm 3, 0)$
$\displaystyle c^2 = a^2 - b^2$
6.13.8
Vertices and Eccentricity of Ellipses
The vertices are the ends of the major axis, $a$ units from the center. Eccentricity measures how stretched the ellipse is: $e$ is between 0 and 1, with 0 a circle and values near 1 long and thin.
e.g. $\tfrac{x^2}{9} + \tfrac{y^2}{25} = 1$ : vertices $(0, \pm 5)$ , $c = 4$ , $e = \tfrac45$
$\displaystyle e = \frac{c}{a}$
6.13.9
Directrices of Ellipses
The two directrices are lines perpendicular to the major axis, $\tfrac{a}{e} = \tfrac{a^2}{c}$ from the center. For any point on the ellipse, its distance to a focus divided by its distance to the matching directrix equals $e$ .
e.g. $\tfrac{x^2}{25} + \tfrac{y^2}{16} = 1$ : $c = 3$ , directrices $x = \pm\tfrac{25}{3}$
$\displaystyle x = h \pm \frac{a^2}{c} \;\text{horizontal}$
$\displaystyle y = k \pm \frac{a^2}{c} \;\text{vertical}$
6.13.10
The Area of an Ellipse
Multiply $\pi$ by the two semi-axes. A circle is the case $a = b = r$ .
e.g. $\tfrac{x^2}{25} + \tfrac{y^2}{9} = 1$ : $A = 15\pi \approx 47.1$
$\displaystyle A = \pi a b$

6.14 Hyperbolas as Conic Sections

A hyperbola opens toward its foci and hugs its dashed asymptotes far from the center. Here c² = a² + b², and from any point on the curve the two focal distances differ by 2a.

6.14.1
Equations of Hyperbolas Centered at the Origin
The positive squared term sets the direction: positive $x^2$ opens left and right, positive $y^2$ opens up and down. $a^2$ always sits under the positive term.
e.g. $\frac{x^2}{9} - \frac{y^2}{16} = 1$ opens left and right with vertices $(\pm 3, 0)$
$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
$\displaystyle \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$
6.14.2
Equations of Hyperbolas Centered at a General Point
Shift the origin form so the center is $(h, k)$ . The signs inside the parentheses are opposite the center's coordinates.
e.g. $\frac{(x-1)^2}{4} - \frac{(y+2)^2}{9} = 1$ : center $(1, -2)$ , vertices $(-1, -2)$ and $(3, -2)$
$\displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$
$\displaystyle \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$
6.14.3
Asymptotes of Hyperbolas Centered at the Origin
Two lines through the center that the branches approach. Shortcut: replace the 1 with 0 and solve for $y$ .
e.g. $\frac{x^2}{9} - \frac{y^2}{16} = 1$ : $y = \pm\tfrac43 x$
$\displaystyle y = \pm\frac{b}{a}x \;\;\text{(opens left/right)}$
$\displaystyle y = \pm\frac{a}{b}x \;\;\text{(opens up/down)}$
6.14.4
Asymptotes of Hyperbolas Centered at a General Point
Same slopes as the origin case, but the lines pass through the center $(h, k)$ .
e.g. $\frac{(x-1)^2}{4} - \frac{(y+2)^2}{9} = 1$ : $y + 2 = \pm\tfrac32(x - 1)$
$\displaystyle y - k = \pm\frac{b}{a}(x - h)$
$\displaystyle \text{(use } \pm\tfrac{a}{b} \text{ if it opens up/down)}$
6.14.5
Finding Intercepts and Intersections of Hyperbolas
Set $y = 0$ for $x$ -intercepts and $x = 0$ for $y$ -intercepts. A negative square means no intercept on that axis. For an intersection with another curve, substitute one equation into the other and solve.
e.g. $x^2 - y^2 = 5$ meets $y = 2$ where $x^2 = 9$ : points $(\pm 3, 2)$
6.14.6
Transverse Axes of Hyperbolas
The segment joining the two vertices. It passes through the center and the foci and points the way the hyperbola opens.
e.g. $\frac{y^2}{25} - \frac{x^2}{4} = 1$ : vertical, from $(0, -5)$ to $(0, 5)$ , length $10$
$\displaystyle \text{length} = 2a$
6.14.7
Conjugate Axes of Hyperbolas
The segment through the center perpendicular to the transverse axis, reaching $b$ units each way. It does not touch the hyperbola.
e.g. $\frac{x^2}{9} - \frac{y^2}{16} = 1$ : from $(0, -4)$ to $(0, 4)$ , length $8$
$\displaystyle \text{length} = 2b$
6.14.8
Foci of Hyperbolas
The foci lie on the transverse axis, $c$ units from the center. For any point on the hyperbola, the difference of its distances to the foci is $2a$ . Note the plus sign, unlike an ellipse.
e.g. $\frac{x^2}{9} - \frac{y^2}{16} = 1$ : $c = \sqrt{25} = 5$ , foci $(\pm 5, 0)$
$\displaystyle c^2 = a^2 + b^2$
6.14.9
Eccentricity and Vertices of Hyperbolas
The vertices are $a$ units from the center along the transverse axis. Eccentricity measures how wide the branches open, and for every hyperbola $e > 1$ .
e.g. $\frac{x^2}{9} - \frac{y^2}{16} = 1$ : vertices $(\pm 3, 0)$ , $e = \tfrac53$
$\displaystyle e = \frac{c}{a} > 1$
$\displaystyle c = \sqrt{a^2 + b^2}$
6.14.10
Directrices of Hyperbolas
Two lines perpendicular to the transverse axis, $\tfrac{a^2}{c}$ units from the center, between the center and each vertex. For any point on the hyperbola, distance to a focus divided by distance to the matching directrix equals $e$ .
e.g. $\frac{x^2}{9} - \frac{y^2}{16} = 1$ : $c = 5$ , directrices $x = \pm\tfrac95$
$\displaystyle x = h \pm \frac{a^2}{c} \;\;\text{(opens left/right)}$
$\displaystyle y = k \pm \frac{a^2}{c} \;\;\text{(opens up/down)}$

Unit 7 10 topics

Trigonometric Functions

Trigonometry with General Triangles, and The Inverse Trigonometric Functions.

7.15 Trigonometry with General Triangles

Each side is named for the angle across from it. These laws work in any triangle, not only right triangles; with C = 90° the law of cosines becomes the Pythagorean theorem.

7.15.1
The Law of Sines
Each side divided by the sine of its opposite angle gives the same ratio. Use it for AAS, ASA, or SSA. SSA is the ambiguous case and can give 0, 1, or 2 triangles.
e.g. $A = 30^\circ$ , $B = 45^\circ$ , $a = 4$ : $b = \frac{4\sin 45^\circ}{\sin 30^\circ} = 4\sqrt2$
$\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
7.15.2
The Law of Cosines
Works for any triangle. Use it for SAS (find the third side) or SSS (find an angle). With $C = 90^\circ$ it becomes the Pythagorean theorem.
e.g. $a = 3$ , $b = 5$ , $C = 120^\circ$ : $c^2 = 9 + 25 - 30(-\tfrac12) = 49$ , so $c = 7$
$\displaystyle c^2 = a^2 + b^2 - 2ab\cos C$
$\displaystyle \cos C = \frac{a^2 + b^2 - c^2}{2ab}$
7.15.3
The Area of a General Triangle
Half the product of two sides times the sine of the angle between them.
e.g. $a = 6$ , $b = 8$ , $C = 30^\circ$ : Area $= \tfrac12(6)(8)(\tfrac12) = 12$
$\displaystyle \text{Area} = \tfrac12 ab\sin C$
7.15.4
Modeling Using the Law of Sines
Sketch the situation and label the triangle. Use the angle sum $180^\circ$ to get the third angle, then pair each side with its opposite angle.
e.g. Stations 10 mi apart sight a fire at $40^\circ$ and $60^\circ$ . The fire's angle is $80^\circ$ , so its distance from the $40^\circ$ station is $\frac{10\sin 60^\circ}{\sin 80^\circ} \approx 8.79$ mi
7.15.5
Modeling Using the Law of Cosines
Use it when a problem gives two distances and the angle between them, or three distances. Make sure the angle you use is the one between the two known sides.
e.g. Hikers leave camp on paths $60^\circ$ apart and walk 3 mi and 8 mi: $d^2 = 9 + 64 - 48\cos 60^\circ = 49$ , so $d = 7$ mi

7.16 The Inverse Trigonometric Functions

Each inverse trig function returns an angle from one fixed arc, so it gives exactly one answer. Arcsine and arctangent use the right half of the circle; arccosine uses the top half, and arctangent never reaches ±π/2.

7.16.1
Graphing the Inverse Sine Function
The graph of $y = \sin^{-1} x$ (also written $\arcsin x$ ) is sine restricted to $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ , reflected over $y = x$ . It increases from $(-1, -\tfrac{\pi}{2})$ through $(0, 0)$ to $(1, \tfrac{\pi}{2})$ .
e.g. $\sin^{-1} 1 = \tfrac{\pi}{2}$ , $\sin^{-1}(-\tfrac12) = -\tfrac{\pi}{6}$
$\displaystyle y = \sin^{-1} x \iff \sin y = x$
$\displaystyle \text{domain } [-1, 1]$
$\displaystyle \text{range } [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$
7.16.2
Graphing the Inverse Cosine Function
The graph of $y = \cos^{-1} x$ is cosine restricted to $[0, \pi]$ , reflected over $y = x$ . It decreases from $(-1, \pi)$ through $(0, \tfrac{\pi}{2})$ to $(1, 0)$ .
e.g. $\cos^{-1} 0 = \tfrac{\pi}{2}$ , $\cos^{-1}(-1) = \pi$
$\displaystyle y = \cos^{-1} x \iff \cos y = x$
$\displaystyle \text{domain } [-1, 1]$
$\displaystyle \text{range } [0, \pi]$
7.16.3
Graphing the Inverse Tangent Function
The graph of $y = \tan^{-1} x$ is tangent restricted to $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ , reflected over $y = x$ . It increases through $(0, 0)$ with horizontal asymptotes $y = \pm\tfrac{\pi}{2}$ .
e.g. $\tan^{-1} 1 = \tfrac{\pi}{4}$ , $\tan^{-1}(-1) = -\tfrac{\pi}{4}$
$\displaystyle y = \tan^{-1} x \iff \tan y = x$
$\displaystyle \text{domain } (-\infty, \infty)$
$\displaystyle \text{range } (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$
7.16.4
Evaluating Expressions Containing Inverse Trigonometric Functions
Find the angle in the inverse function's range that has the given value. A trig function undoes its own inverse whenever the input is in the inverse's domain.
e.g. $\cos^{-1}(-\tfrac{\sqrt2}{2}) = \tfrac{3\pi}{4}$
$\displaystyle \sin(\sin^{-1} x) = x,\; -1 \le x \le 1$
$\displaystyle \cos(\cos^{-1} x) = x,\; -1 \le x \le 1$
$\displaystyle \tan(\tan^{-1} x) = x \text{ for all } x$
7.16.5
Further Evaluating Expressions Containing Inverse Trigonometric Functions
For a trig function of an inverse trig value, draw a right triangle from the input and read off the ratio, choosing the sign by quadrant. For an inverse of a trig value, the answer must land in the inverse's range.
e.g. $\cos(\sin^{-1}\tfrac35) = \tfrac45$ (3-4-5 triangle); $\sin^{-1}(\sin\tfrac{5\pi}{6}) = \tfrac{\pi}{6}$
$\displaystyle \sin^{-1}(\sin\theta) = \theta$
$\displaystyle \text{only if } -\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$

Unit 8 22 topics

Trigonometric Identities

Trigonometric Identities, The Sum and Difference Formulas, and The Double-Angle Formulas.

8.17 Trigonometric Identities

On the unit circle, the point at angle θ is (cos θ, sin θ), so the legs of the triangle are cos θ and sin θ. The quadrant labels show which ratios are positive there.

8.17.1
Simplifying Expressions Using Basic Trigonometric Identities
Rewrite everything in sine and cosine using the quotient and reciprocal identities, then cancel.
e.g. $\tan\theta\cos\theta = \tfrac{\sin\theta}{\cos\theta}\cdot\cos\theta = \sin\theta$
$\displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta},\; \cot\theta = \frac{\cos\theta}{\sin\theta}$
$\displaystyle \sec\theta = \frac{1}{\cos\theta},\; \csc\theta = \frac{1}{\sin\theta}$
8.17.2
Simplifying Expressions Using the Pythagorean Identity
Holds for every angle. Look for $\sin^2\theta + \cos^2\theta$ and replace it with 1.
e.g. $\frac{1 - \sin^2\theta}{\cos\theta} = \frac{\cos^2\theta}{\cos\theta} = \cos\theta$
$\displaystyle \sin^2\theta + \cos^2\theta = 1$
8.17.3
Alternate Forms of the Pythagorean Identity
Solve the identity for one squared term. When taking a square root, pick the sign from the angle's quadrant.
e.g. $\sin\theta = \tfrac35$ in Quadrant II: $\cos\theta = -\sqrt{1 - \tfrac{9}{25}} = -\tfrac45$
$\displaystyle \sin^2\theta = 1 - \cos^2\theta$
$\displaystyle \cos^2\theta = 1 - \sin^2\theta$
$\displaystyle \cos\theta = \pm\sqrt{1 - \sin^2\theta}$
8.17.4
Simplifying Expressions Using the Secant-Tangent Identity
Divide the Pythagorean identity by $\cos^2\theta$ to get it. Look for $1 + \tan^2\theta$ or $\sec^2\theta - \tan^2\theta$ .
e.g. $(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = \sec^2\theta - \tan^2\theta = 1$
$\displaystyle 1 + \tan^2\theta = \sec^2\theta$
8.17.5
Alternate Forms of the Secant-Tangent Identity
Rearrange $1 + \tan^2\theta = \sec^2\theta$ as needed. Choose the sign of a square root from the quadrant.
e.g. $\tan\theta = \tfrac34$ in Quadrant III: $\sec\theta = -\sqrt{1 + \tfrac{9}{16}} = -\tfrac54$
$\displaystyle \tan^2\theta = \sec^2\theta - 1$
$\displaystyle \sec^2\theta - \tan^2\theta = 1$
$\displaystyle \sec\theta = \pm\sqrt{1 + \tan^2\theta}$
8.17.6
Simplifying Trigonometric Expressions Using the Cotangent-Cosecant Identity
Divide the Pythagorean identity by $\sin^2\theta$ to get it. Rearranged forms are $\cot^2\theta = \csc^2\theta - 1$ and $\csc^2\theta - \cot^2\theta = 1$ .
e.g. $\sin^2\theta(1 + \cot^2\theta) = \sin^2\theta\csc^2\theta = 1$
$\displaystyle 1 + \cot^2\theta = \csc^2\theta$
8.17.7
Simplifying Trigonometric Expressions Using Cofunction Identities
A function of an angle equals its cofunction of the complementary angle. The pairs are sine and cosine, tangent and cotangent, secant and cosecant. In degrees, use $90^\circ - \theta$ .
e.g. $\cos(\tfrac{\pi}{2} - \theta)\csc\theta = \sin\theta\csc\theta = 1$
$\displaystyle \sin(\tfrac{\pi}{2} - \theta) = \cos\theta$
$\displaystyle \cos(\tfrac{\pi}{2} - \theta) = \sin\theta$
$\displaystyle \tan(\tfrac{\pi}{2} - \theta) = \cot\theta$
$\displaystyle \sec(\tfrac{\pi}{2} - \theta) = \csc\theta$

8.18 The Sum and Difference Formulas

8.18.1
The Sum and Difference Formulas for Sine
The signs match: plus on the left gives plus on the right. Note $\sin(A + B) \ne \sin A + \sin B$ .
e.g. $\sin(x + \pi) = \sin x\cos\pi + \cos x\sin\pi = -\sin x$
$\displaystyle \sin(A + B) = \sin A\cos B + \cos A\sin B$
$\displaystyle \sin(A - B) = \sin A\cos B - \cos A\sin B$
8.18.2
The Sum and Difference Formulas for Cosine
The sign flips: a sum on the left gives a minus on the right.
e.g. $\cos(x + \tfrac{\pi}{2}) = \cos x\cdot 0 - \sin x\cdot 1 = -\sin x$
$\displaystyle \cos(A + B) = \cos A\cos B - \sin A\sin B$
$\displaystyle \cos(A - B) = \cos A\cos B + \sin A\sin B$
8.18.3
The Sum and Difference Formulas for Tangent
The numerator keeps the sign and the denominator flips it. Undefined when the denominator is 0.
e.g. $\tan(x + \tfrac{\pi}{4}) = \frac{\tan x + 1}{1 - \tan x}$
$\displaystyle \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$
$\displaystyle \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$
8.18.4
Calculating Trigonometric Ratios Using the Sum Formula for Sine
Write the angle as a sum or difference of special angles ( $30^\circ$ , $45^\circ$ , $60^\circ$ , or their radian forms), then apply the formula. If ratios are given instead, find any missing ones with the Pythagorean identity.
e.g. $\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \frac{\sqrt6 + \sqrt2}{4}$
$\displaystyle \sin(A + B) = \sin A\cos B + \cos A\sin B$
8.18.5
Calculating Trigonometric Ratios Using the Sum Formula for Cosine
Split the angle into special angles, or find the missing ratios from given ones, then apply the formula. Watch the minus sign.
e.g. $\cos 75^\circ = \cos 45^\circ\cos 30^\circ - \sin 45^\circ\sin 30^\circ = \frac{\sqrt6 - \sqrt2}{4}$
$\displaystyle \cos(A + B) = \cos A\cos B - \sin A\sin B$
8.18.6
Calculating Trigonometric Ratios Using the Sum Formula for Tangent
Split the angle into special angles with known tangents ( $\tan 30^\circ = \tfrac{\sqrt3}{3}$ , $\tan 45^\circ = 1$ , $\tan 60^\circ = \sqrt3$ ), apply the formula, and rationalize.
e.g. $\tan 75^\circ = \frac{1 + \sqrt3/3}{1 - \sqrt3/3} = \frac{3 + \sqrt3}{3 - \sqrt3} = 2 + \sqrt3$
$\displaystyle \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$
8.18.7
Writing Sums of Trigonometric Functions in Amplitude-Phase Form
A sum $a\sin x + b\cos x$ is a single sine wave with amplitude $R$ . Find the phase $\varphi$ from its cosine and sine, using both signs to fix the quadrant.
e.g. $\sin x + \sqrt3\cos x$ : $R = 2$ , $\varphi = \tfrac{\pi}{3}$ , so it equals $2\sin(x + \tfrac{\pi}{3})$
$\displaystyle a\sin x + b\cos x = R\sin(x + \varphi)$
$\displaystyle R = \sqrt{a^2 + b^2}$
$\displaystyle \cos\varphi = \tfrac{a}{R},\; \sin\varphi = \tfrac{b}{R}$

8.19 The Double-Angle Formulas

8.19.1
The Double-Angle Formula for Sine
The sine of twice an angle. It comes from the sum formula with $A = B = \theta$ . Note $\sin 2\theta \ne 2\sin\theta$ .
e.g. $2\sin 15^\circ\cos 15^\circ = \sin 30^\circ = \tfrac12$
$\displaystyle \sin 2\theta = 2\sin\theta\cos\theta$
8.19.2
Verifying Trigonometric Identities Using the Double-Angle Formula for Sine
Start with the more complicated side and transform it into the other side. Replace $\sin 2\theta$ with $2\sin\theta\cos\theta$ , or spot $2\sin\theta\cos\theta$ and collapse it.
e.g. $(\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 + \sin 2\theta$
$\displaystyle \sin 2\theta = 2\sin\theta\cos\theta$
8.19.3
Using the Double-Angle Formula for Sine With the Pythagorean Theorem
Given one ratio and the quadrant, draw a right triangle (or use the Pythagorean identity) to find the other ratio. Give it the sign from the quadrant, then double.
e.g. $\cos\theta = \tfrac{5}{13}$ in Quadrant IV: $\sin\theta = -\tfrac{12}{13}$ , so $\sin 2\theta = 2(-\tfrac{12}{13})(\tfrac{5}{13}) = -\tfrac{120}{169}$
$\displaystyle \sin 2\theta = 2\sin\theta\cos\theta$
8.19.4
The Double-Angle Formula for Cosine
Three equivalent forms. Pick the one that uses the ratio you know.
e.g. $\sin\theta = \tfrac13$ : $\cos 2\theta = 1 - 2\cdot\tfrac19 = \tfrac79$
$\displaystyle \cos 2\theta = \cos^2\theta - \sin^2\theta$
$\displaystyle = 2\cos^2\theta - 1$
$\displaystyle = 1 - 2\sin^2\theta$
8.19.5
Verifying Trigonometric Identities Using the Double-Angle Formulas for Cosine
Choose the form that cancels the constant: with $1 - \cos 2\theta$ use $1 - 2\sin^2\theta$ , and with $1 + \cos 2\theta$ use $2\cos^2\theta - 1$ .
e.g. $\frac{1 - \cos 2\theta}{\sin 2\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \tan\theta$
$\displaystyle 1 - \cos 2\theta = 2\sin^2\theta$
$\displaystyle 1 + \cos 2\theta = 2\cos^2\theta$
8.19.6
Finding Exact Values of Trigonometric Expressions Using the Double-Angle Formulas for
Spot a double-angle pattern and collapse it to one function of the doubled angle, then use a known exact value.
e.g. $\cos^2 15^\circ - \sin^2 15^\circ = \cos 30^\circ = \tfrac{\sqrt3}{2}$
$\displaystyle \cos^2\theta - \sin^2\theta = \cos 2\theta$
$\displaystyle 2\cos^2\theta - 1 = \cos 2\theta$
$\displaystyle 1 - 2\sin^2\theta = \cos 2\theta$
8.19.7
Simplifying Expressions Using the Double-Angle Formula for Tangent
Spot the pattern $\frac{2\tan\theta}{1 - \tan^2\theta}$ and collapse it to $\tan 2\theta$ . Undefined when $\tan^2\theta = 1$ .
e.g. $\frac{2\tan 15^\circ}{1 - \tan^2 15^\circ} = \tan 30^\circ = \tfrac{\sqrt3}{3}$
$\displaystyle \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$
8.19.8
Verifying Trigonometric Identities Using the Double-Angle Formula for Tangent
Replace $\tan 2\theta$ with its formula, then simplify toward the other side. Multiplying top and bottom by the same expression often helps.
e.g. $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$ ; multiply top and bottom by $\cot\theta$ to get $\frac{2}{\cot\theta - \tan\theta}$
$\displaystyle \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$

Unit 9 15 topics

Trigonometric Equations

Elementary Trigonometric Equations, Trigonometric Equations Containing Transformed Functions, and Advanced Trigonometric Equations.

9.20 Elementary Trigonometric Equations

The line y = ½ meets the unit circle at π/6 and 5π/6, and it meets the sine wave at the same two x-values. Adding 2πn to each gives every solution.

9.20.1
Elementary Trigonometric Equations Containing Sine
Isolate $\sin x$ . If $\sin x = k$ with $-1 \le k \le 1$ , the solutions in $[0, 2\pi)$ are $\alpha$ and $\pi - \alpha$ , where $\alpha = \sin^{-1} k$ . Add $2\pi$ to any negative angle, and there is no solution if $\lvert k \rvert > 1$ .
e.g. $\sin x = \tfrac12 \Rightarrow x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$
$\displaystyle \sin x = k \Rightarrow x = \alpha,\; \pi - \alpha$
$\displaystyle \alpha = \sin^{-1} k$
9.20.2
Elementary Trigonometric Equations Containing Cosine
Isolate $\cos x$ . If $\cos x = k$ with $-1 \le k \le 1$ , the solutions in $[0, 2\pi)$ are $\alpha$ and $2\pi - \alpha$ , where $\alpha = \cos^{-1} k$ . There is no solution if $\lvert k \rvert > 1$ .
e.g. $\cos x = -\tfrac12 \Rightarrow x = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}$
$\displaystyle \cos x = k \Rightarrow x = \alpha,\; 2\pi - \alpha$
$\displaystyle \alpha = \cos^{-1} k$
9.20.3
Elementary Trigonometric Equations Containing Tangent
$\tan x = k$ has solutions for every real $k$ . In $[0, 2\pi)$ they are $\alpha$ and $\alpha + \pi$ , where $\alpha = \tan^{-1} k$ . If $\alpha$ is negative, use $\alpha + \pi$ and $\alpha + 2\pi$ .
e.g. $\tan x = -1$ : $\alpha = -\tfrac{\pi}{4}$ , so $x = \tfrac{3\pi}{4}, \tfrac{7\pi}{4}$
$\displaystyle \tan x = k \Rightarrow x = \alpha,\; \alpha + \pi$
$\displaystyle \alpha = \tan^{-1} k$
9.20.4
Elementary Trigonometric Equations Containing Secant
Rewrite as a cosine equation and solve that. There is no solution if $\lvert k \rvert < 1$ .
e.g. $\sec x = 2 \Rightarrow \cos x = \tfrac12 \Rightarrow x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$
$\displaystyle \sec x = k \iff \cos x = \frac{1}{k}$
9.20.5
Elementary Trigonometric Equations Containing Cosecant
Rewrite as a sine equation and solve that. There is no solution if $\lvert k \rvert < 1$ .
e.g. $\csc x = -2 \Rightarrow \sin x = -\tfrac12 \Rightarrow x = \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$
$\displaystyle \csc x = k \iff \sin x = \frac{1}{k}$
9.20.6
Elementary Trigonometric Equations Containing Cotangent
For $k \ne 0$ , rewrite as a tangent equation. For $k = 0$ , use $\cos x = 0$ , so $x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}$ .
e.g. $\cot x = \sqrt3 \Rightarrow \tan x = \tfrac{1}{\sqrt3} \Rightarrow x = \tfrac{\pi}{6}, \tfrac{7\pi}{6}$
$\displaystyle \cot x = k \iff \tan x = \frac{1}{k}$
9.20.7
General Solutions of Elementary Trigonometric Equations
Find one solution $\alpha$ , then add every whole number of periods. Sine and cosine repeat every $2\pi$ , tangent every $\pi$ . Here $n$ is any integer.
e.g. $\cos x = \tfrac12 \Rightarrow x = \pm\tfrac{\pi}{3} + 2\pi n$
$\displaystyle \sin x = k:\; x = \alpha + 2\pi n$
$\displaystyle \text{or } x = \pi - \alpha + 2\pi n$
$\displaystyle \cos x = k:\; x = \pm\alpha + 2\pi n$
$\displaystyle \tan x = k:\; x = \alpha + \pi n$

9.21 Trigonometric Equations Containing Transformed Functions

A sinusoid is the sine wave stretched, shifted, and lifted. Read the midline first; the amplitude is how far the wave rises above it, and the period is the length of one full cycle.

9.21.1
General Solutions of Trigonometric Equations With Transformed Functions
Let $u = bx + c$ . Write the general solution for $u$ , then solve for $x$ . Dividing by $b$ also divides the period, so $2\pi n$ becomes $\tfrac{2\pi n}{b}$ .
e.g. $\sin 2x = \tfrac12 \Rightarrow 2x = \tfrac{\pi}{6} + 2\pi n$ or $\tfrac{5\pi}{6} + 2\pi n$ , so $x = \tfrac{\pi}{12} + \pi n$ or $\tfrac{5\pi}{12} + \pi n$
$\displaystyle \sin(bx + c) = k$
$\displaystyle \Rightarrow\; bx + c = \alpha + 2\pi n$
$\displaystyle \text{or } bx + c = \pi - \alpha + 2\pi n$
9.21.2
Trigonometric Equations Containing Transformed Sine Functions
Isolate $\sin(bx + c)$ and let $u = bx + c$ . If $0 \le x < 2\pi$ , then $u$ runs from $c$ to $2\pi b + c$ , so list every solution for $u$ in that longer interval, then solve for $x$ .
e.g. $2\sin 2x = \sqrt3$ on $[0, 2\pi)$ : $2x = \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \tfrac{7\pi}{3}, \tfrac{8\pi}{3}$ , so $x = \tfrac{\pi}{6}, \tfrac{\pi}{3}, \tfrac{7\pi}{6}, \tfrac{4\pi}{3}$
$\displaystyle a\sin(bx + c) + d = k$
$\displaystyle \Rightarrow\; \sin u = \frac{k - d}{a}$
9.21.3
Trigonometric Equations Containing Transformed Cosine Functions
Isolate $\cos(bx + c)$ and let $u = bx + c$ . Shift the interval for $x$ to get the interval for $u$ , find every $u$ in it, then solve for $x$ .
e.g. $\cos(x - \tfrac{\pi}{4}) = 0$ on $[0, 2\pi)$ : $u \in [-\tfrac{\pi}{4}, \tfrac{7\pi}{4})$ , $u = \tfrac{\pi}{2}, \tfrac{3\pi}{2}$ , so $x = \tfrac{3\pi}{4}, \tfrac{7\pi}{4}$
$\displaystyle a\cos(bx + c) + d = k$
$\displaystyle \Rightarrow\; \cos u = \frac{k - d}{a}$
9.21.4
Trigonometric Equations Containing Transformed Tangent Functions
Let $u = bx + c$ and solve $\tan u = k$ with period $\pi$ . Then solve for $x$ , so the period becomes $\tfrac{\pi}{b}$ .
e.g. $\tan 2x = 1 \Rightarrow 2x = \tfrac{\pi}{4} + \pi n \Rightarrow x = \tfrac{\pi}{8} + \tfrac{\pi n}{2}$
$\displaystyle \tan(bx + c) = k$
$\displaystyle \Rightarrow\; bx + c = \alpha + \pi n$

9.22 Advanced Trigonometric Equations

9.22.1
Solving Trigonometric Equations Using the Sin-Cos-Tan Identity
When an equation has the form $\sin x = k\cos x$ , divide both sides by $\cos x$ to get $\tan x = k$ . First check that $\cos x = 0$ is not a solution, since those values would be lost.
e.g. $\sin x = \sqrt3\cos x \Rightarrow \tan x = \sqrt3 \Rightarrow x = \tfrac{\pi}{3}, \tfrac{4\pi}{3}$
$\displaystyle \tan x = \frac{\sin x}{\cos x}$
9.22.2
Solving Trigonometric Equations Using the Zero-Product Property
Move everything to one side, factor, and set each factor equal to zero. Never divide both sides by a trig function, because that throws away solutions.
e.g. $2\sin x\cos x = \cos x \Rightarrow \cos x(2\sin x - 1) = 0$ , so $x = \tfrac{\pi}{6}, \tfrac{\pi}{2}, \tfrac{5\pi}{6}, \tfrac{3\pi}{2}$
$\displaystyle AB = 0 \Rightarrow A = 0 \;\text{or}\; B = 0$
9.22.3
Quadratic Trigonometric Equations Containing Sine or Cosine
Use $\sin^2 x + \cos^2 x = 1$ to write everything in one function. Treat it as a quadratic in $u = \sin x$ (or $\cos x$ ), factor or use the quadratic formula, and reject any root outside $[-1, 1]$ .
e.g. $2\sin^2 x - \sin x - 1 = 0 \Rightarrow (2\sin x + 1)(\sin x - 1) = 0$ , so $x = \tfrac{\pi}{2}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$
$\displaystyle \sin^2 x + \cos^2 x = 1$
9.22.4
Quadratic Trigonometric Equations Containing Tangent or Cotangent
Rewrite in terms of $\tan x$ alone, then solve the quadratic in $u = \tan x$ . Tangent takes every real value, so no root is rejected for size, but check that no step divided by zero.
e.g. $\tan x + \cot x = 2 \Rightarrow \tan^2 x - 2\tan x + 1 = 0 \Rightarrow \tan x = 1$ , so $x = \tfrac{\pi}{4}, \tfrac{5\pi}{4}$
$\displaystyle \cot x = \frac{1}{\tan x}$
$\displaystyle 1 + \tan^2 x = \sec^2 x$

Unit 10 10 topics

Parametric Equations

Parametric Equations.

10.23 Parametric Equations

A parametric curve gives x and y separately as functions of t. Plotting points in order of t shows the path and its direction, which an ordinary y = f(x) graph cannot show.

10.23.1
Graphing Curves Defined Parametrically
Both $x$ and $y$ are given in terms of a parameter $t$ . Make a table of $t$ values, plot the points $(x, y)$ , and connect them in order of increasing $t$ . Arrows show the direction (orientation).
e.g. $x = t + 1$ , $y = 2t$ : $t = 0, 1, 2$ gives $(1, 0), (2, 2), (3, 4)$
$\displaystyle x = f(t), \quad y = g(t)$
10.23.2
Cartesian Equations of Parametric Curves
Eliminate the parameter. Solve one equation for $t$ and substitute into the other, or use an identity when $t$ sits inside sine and cosine. Keep any restriction the parameter places on $x$ or $y$ .
e.g. $x = t + 1$ , $y = t^2 \Rightarrow t = x - 1 \Rightarrow y = (x - 1)^2$
$\displaystyle \cos^2 t + \sin^2 t = 1$
10.23.3
Finding Intercepts of Curves Defined Parametrically
For the $y$ -intercept, set $x(t) = 0$ , solve for $t$ , and find $y$ . For the $x$ -intercept, set $y(t) = 0$ , solve for $t$ , and find $x$ .
e.g. $x = t - 2$ , $y = t^2 - 1$ : $x = 0$ at $t = 2$ gives $(0, 3)$ ; $y = 0$ at $t = \pm1$ gives $(-1, 0), (-3, 0)$
$\displaystyle x(t) = 0 \to y\text{-intercept}$
$\displaystyle y(t) = 0 \to x\text{-intercept}$
10.23.4
Finding Intersections of Parametric Curves and Lines
Substitute $x(t)$ and $y(t)$ into the equation of the line, solve for $t$ , and plug each $t$ back in to get the points.
e.g. $x = t$ , $y = t^2$ , line $y = x + 2$ : $t^2 - t - 2 = 0 \Rightarrow t = 2, -1$ , points $(2, 4), (-1, 1)$
10.23.5
Parametric Equations of Circles
A circle with center $(h, k)$ and radius $r$ , traced counterclockwise once for $0 \le t < 2\pi$ , starting at $(h + r, k)$ .
e.g. $x = 1 + 2\cos t$ , $y = -2 + 2\sin t$ : center $(1, -2)$ , radius $2$
$\displaystyle x = h + r\cos t$
$\displaystyle y = k + r\sin t$
$\displaystyle (x - h)^2 + (y - k)^2 = r^2$
10.23.6
Parametric Equations of Ellipses
An ellipse with center $(h, k)$ , horizontal semi-axis $a$ , and vertical semi-axis $b$ .
e.g. $x = 4\cos t$ , $y = 2\sin t \Rightarrow \tfrac{x^2}{16} + \tfrac{y^2}{4} = 1$
$\displaystyle x = h + a\cos t$
$\displaystyle y = k + b\sin t$
$\displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$
10.23.7
Parametric Equations of Parabolas
A parabola with vertex at the origin. For $y^2 = 4ax$ (opening sideways) use $x = at^2$ , $y = 2at$ . For $x^2 = 4ay$ (opening up or down) use $x = 2at$ , $y = at^2$ .
e.g. $y^2 = 12x$ : $a = 3$ , so $x = 3t^2$ , $y = 6t$
$\displaystyle y^2 = 4ax:\; x = at^2,\; y = 2at$
$\displaystyle x^2 = 4ay:\; x = 2at,\; y = at^2$
10.23.8
Parametric Equations of Parabolas Centered at (h,k)
Shift the vertex to $(h, k)$ by adding $h$ to $x$ and $k$ to $y$ .
e.g. $(y - 2)^2 = 8(x + 1)$ : $a = 2$ , so $x = -1 + 2t^2$ , $y = 2 + 4t$
$\displaystyle (y - k)^2 = 4a(x - h)$
$\displaystyle \Rightarrow\; x = h + at^2,\; y = k + 2at$
$\displaystyle (x - h)^2 = 4a(y - k)$
$\displaystyle \Rightarrow\; x = h + 2at,\; y = k + at^2$
10.23.9
Parametric Equations of Horizontal Hyperbolas
A hyperbola opening left and right, centered at $(h, k)$ . It works because $\sec^2 t - \tan^2 t = 1$ .
e.g. $x = 3\sec t$ , $y = 2\tan t \Rightarrow \tfrac{x^2}{9} - \tfrac{y^2}{4} = 1$
$\displaystyle x = h + a\sec t$
$\displaystyle y = k + b\tan t$
$\displaystyle \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$
10.23.10
Parametric Equations of Vertical Hyperbolas
A hyperbola opening up and down, centered at $(h, k)$ . Secant goes with $y$ and tangent goes with $x$ .
e.g. $x = 4\tan t$ , $y = 5\sec t \Rightarrow \tfrac{y^2}{25} - \tfrac{x^2}{16} = 1$
$\displaystyle x = h + b\tan t$
$\displaystyle y = k + a\sec t$
$\displaystyle \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$

Unit 11 6 topics

Polar Equations

Polar Coordinates.

11.24 Polar Coordinates

A polar point is a distance r from the pole at an angle θ. The conversion formulas come straight from the right triangle with legs x and y and hypotenuse r.

11.24.1
Introduction to Polar Coordinates
A point $(r, \theta)$ is $r$ units from the origin (the pole) at angle $\theta$ measured counterclockwise from the positive $x$ -axis. A negative $r$ points the opposite way. Every point has infinitely many polar names.
e.g. $(2, \tfrac{\pi}{3}) = (2, \tfrac{7\pi}{3}) = (-2, \tfrac{4\pi}{3})$
$\displaystyle (r, \theta) = (r, \theta + 2\pi n)$
$\displaystyle (r, \theta) = (-r, \theta + \pi)$
11.24.2
Converting from Polar Coordinates to Cartesian Coordinates
Use cosine for $x$ and sine for $y$ . To convert an equation, substitute these and use $r^2 = x^2 + y^2$ .
e.g. $(4, \tfrac{\pi}{6}) \to (4 \cdot \tfrac{\sqrt3}{2}, 4 \cdot \tfrac12) = (2\sqrt3, 2)$
$\displaystyle x = r\cos\theta$
$\displaystyle y = r\sin\theta$
$\displaystyle r^2 = x^2 + y^2$
11.24.3
Polar Equations of Circles Centered at the Origin
$r = a$ is a circle centered at the origin with radius $\lvert a \rvert$ .
e.g. $r = 3$ : circle $x^2 + y^2 = 9$
$\displaystyle r = a \iff x^2 + y^2 = a^2$
11.24.4
Polar Equations of Radial Lines
$\theta = \alpha$ is the whole line through the origin at angle $\alpha$ , because $r$ can be positive or negative.
e.g. $\theta = \tfrac{\pi}{4}$ is $y = x$ . $\theta = \tfrac{\pi}{2}$ is $x = 0$ .
$\displaystyle \theta = \alpha \iff y = (\tan\alpha)\, x$
11.24.5
Polar Equations of Circles Centered on the Coordinate Axes
These circles pass through the origin and have radius $\lvert a \rvert$ . Cosine puts the center on the $x$ -axis, sine puts it on the $y$ -axis.
e.g. $r = 6\sin\theta \Rightarrow x^2 + y^2 = 6y \Rightarrow x^2 + (y - 3)^2 = 9$ , center $(0, 3)$ , radius $3$
$\displaystyle r = 2a\cos\theta:\; \text{center } (a, 0)$
$\displaystyle r = 2a\sin\theta:\; \text{center } (0, a)$
11.24.6
Finding Intersections of Polar Curves
Set the two expressions for $r$ equal and solve for $\theta$ . Check the pole separately, since both curves can pass through it at different angles.
e.g. $r = \sin\theta$ and $r = \cos\theta$ : $\tan\theta = 1$ gives $(\tfrac{\sqrt2}{2}, \tfrac{\pi}{4})$ , plus the pole
$\displaystyle r_1(\theta) = r_2(\theta)$

Unit 12 28 topics

Complex Numbers

The Complex Plane, Further Complex Numbers, Complex Numbers in Polar Form, and more.

12.25 The Complex Plane

The number 3 + 2i sits 3 units along the real axis and 2 units up the imaginary axis. Each multiplication by i turns a quarter turn, so the powers of i cycle i, −1, −i, 1.

12.25.1
The Complex Plane
The complex number $a + bi$ is plotted as the point $(a, b)$ . The horizontal axis is the real axis and the vertical axis is the imaginary axis.
e.g. $3 - 2i$ is the point $(3, -2)$
$\displaystyle a + bi \;\to\; (a, b)$
12.25.2
The Magnitude of a Complex Number
The magnitude (modulus) $\lvert z \rvert$ is the distance from $z$ to the origin. It is always a nonnegative real number.
e.g. $\lvert 3 + 4i \rvert = \sqrt{9 + 16} = 5$
$\displaystyle \lvert a + bi \rvert = \sqrt{a^2 + b^2}$
12.25.3
The Argument of a Complex Number
The argument $\arg z$ is the angle from the positive real axis to $z$ . Use $\tan\theta = \tfrac{b}{a}$ , then pick the angle in the correct quadrant. The principal argument satisfies $-\pi < \theta \le \pi$ .
e.g. $\arg(-1 + i) = \tfrac{3\pi}{4}$ , $\arg(1 - i) = -\tfrac{\pi}{4}$
$\displaystyle \tan\theta = \frac{b}{a}$
$\displaystyle z = \lvert z \rvert(\cos\theta + i\sin\theta)$
12.25.4
Arithmetic in the Complex Plane
Adding complex numbers adds them like vectors (parallelogram rule), and subtracting works the same way. Multiplying by a positive real $k$ scales the distance from the origin by $k$ . Multiplying by $i$ rotates $90^\circ$ counterclockwise.
e.g. $(2 + i) + (1 + 3i) = 3 + 4i$ ; $i(2 + i) = -1 + 2i$ , a $90^\circ$ turn
$\displaystyle (a + bi) + (c + di)$
$\displaystyle = (a + c) + (b + d)i$
12.25.5
Geometry in the Complex Plane
The distance between $z_1$ and $z_2$ is $\lvert z_1 - z_2 \rvert$ . The midpoint is their average. The points with $\lvert z - z_0 \rvert = r$ form a circle with center $z_0$ and radius $r$ .
e.g. Distance from $1 + i$ to $4 + 5i$ : $\lvert 3 + 4i \rvert = 5$
$\displaystyle d = \lvert z_1 - z_2 \rvert$
$\displaystyle M = \frac{z_1 + z_2}{2}$

12.26 Further Complex Numbers

12.26.1
The Complex Conjugate
The conjugate of $z = a + bi$ is $\bar z = a - bi$ : flip the sign of the imaginary part. In the complex plane it is the reflection of $z$ across the real axis.
e.g. $\overline{-2 + 5i} = -2 - 5i$
$\displaystyle \overline{a + bi} = a - bi$
12.26.2
Special Properties of the Complex Conjugate
A number times its conjugate is a nonnegative real number. Conjugation passes through sums and products.
e.g. $(3 + 2i)(3 - 2i) = 9 + 4 = 13$
$\displaystyle z\bar z = \lvert z \rvert^2 = a^2 + b^2$
$\displaystyle z + \bar z = 2a,\quad z - \bar z = 2bi$
$\displaystyle \overline{z + w} = \bar z + \bar w$
$\displaystyle \overline{zw} = \bar z\,\bar w$
12.26.3
The Complex Conjugate and the Roots of a Quadratic Equation
If a quadratic with real coefficients has a negative discriminant, its roots are a conjugate pair $p \pm qi$ . Their sum is $2p$ and their product is $p^2 + q^2$ .
e.g. $x^2 - 4x + 13 = 0 \Rightarrow x = \tfrac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i$
$\displaystyle \text{roots } p \pm qi:$
$\displaystyle x^2 - 2px + (p^2 + q^2) = 0$
12.26.4
Dividing Complex Numbers
Multiply the top and bottom by the conjugate of the denominator. The denominator becomes the real number $c^2 + d^2$ .
e.g. $\tfrac{5 + 5i}{1 + 2i} = \tfrac{(5 + 5i)(1 - 2i)}{5} = \tfrac{15 - 5i}{5} = 3 - i$
$\displaystyle \frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{c^2 + d^2}$
12.26.5
Solving Equations by Equating Real and Imaginary Parts
Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal. This turns one complex equation into two real equations.
e.g. $(x + y) + (x - y)i = 5 + i \Rightarrow x + y = 5$ , $x - y = 1$ , so $x = 3$ , $y = 2$
$\displaystyle a + bi = c + di \iff a = c,\; b = d$
12.26.6
Extending Polynomial Identities to the Complex Numbers
The usual identities still hold with complex numbers. With $i$ , a sum of squares factors like a difference of squares.
e.g. $x^2 + 9 = (x + 3i)(x - 3i)$ , so the roots are $\pm 3i$
$\displaystyle a^2 + b^2 = (a + bi)(a - bi)$
$\displaystyle (a + b)^2 = a^2 + 2ab + b^2$

12.27 Complex Numbers in Polar Form

A complex number a + bi is a point in the plane, so it also has a length r and an angle θ. In that polar form the n-th roots of 1 sit evenly around the unit circle, like the corners of a regular hexagon.

12.27.1
The Polar Form of a Complex Number
Write $z = a + bi$ using its distance from the origin and its angle. The modulus $r = \lvert z \rvert$ is the length, and the argument $\theta$ is measured counterclockwise from the positive real axis. Use the signs of $a$ and $b$ to put $\theta$ in the right quadrant.
e.g. $1 + i$ : $r = \sqrt2$ , $\theta = \tfrac{\pi}{4}$ , so $z = \sqrt2\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right)$
$\displaystyle z = r(\cos\theta + i\sin\theta)$
$\displaystyle r = \sqrt{a^2 + b^2}$
$\displaystyle \tan\theta = \frac{b}{a}$
12.27.2
Products of Complex Numbers Expressed in Polar Form
To multiply, multiply the moduli and add the arguments.
e.g. $2(\cos 30^\circ + i\sin 30^\circ) \cdot 3(\cos 60^\circ + i\sin 60^\circ) = 6(\cos 90^\circ + i\sin 90^\circ) = 6i$
$\displaystyle \lvert z_1 z_2 \rvert = r_1 r_2$
$\displaystyle \arg(z_1 z_2) = \theta_1 + \theta_2$
12.27.3
Quotients of Complex Numbers Expressed in Polar Form
To divide, divide the moduli and subtract the arguments. The divisor cannot be $0$ .
e.g. $\dfrac{8(\cos 100^\circ + i\sin 100^\circ)}{2(\cos 10^\circ + i\sin 10^\circ)} = 4(\cos 90^\circ + i\sin 90^\circ) = 4i$
$\displaystyle \left\lvert \frac{z_1}{z_2} \right\rvert = \frac{r_1}{r_2}$
$\displaystyle \arg\frac{z_1}{z_2} = \theta_1 - \theta_2$
12.27.4
The CIS Notation
$\operatorname{cis}\theta$ is shorthand for $\cos\theta + i\sin\theta$ . Products and quotients become $r_1 r_2 \operatorname{cis}(\theta_1 + \theta_2)$ and $\tfrac{r_1}{r_2}\operatorname{cis}(\theta_1 - \theta_2)$ .
e.g. $3\operatorname{cis}\tfrac{\pi}{2} = 3(0 + 1i) = 3i$
$\displaystyle \operatorname{cis}\theta = \cos\theta + i\sin\theta$
$\displaystyle z = r\operatorname{cis}\theta$

12.28 De Moivre's Theorem

12.28.1
De Moivre's Theorem
To raise a complex number in polar form to an integer power $n$ , raise the modulus to the $n$ th power and multiply the argument by $n$ .
e.g. $\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right)^3 = \cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2} = i$
$\displaystyle [r(\cos\theta + i\sin\theta)]^n$
$\displaystyle = r^n(\cos n\theta + i\sin n\theta)$
12.28.2
Finding Powers of Complex Numbers Using De Moivre's Theorem
Convert $a + bi$ to polar form, apply De Moivre's theorem, then convert back to $a + bi$ if asked. Reduce the final angle by multiples of $2\pi$ (or $360^\circ$ ).
e.g. $(1 + i)^8$ : $r = \sqrt2$ , $\theta = \tfrac{\pi}{4}$ , so $(\sqrt2)^8 \operatorname{cis} 2\pi = 16$
$\displaystyle z^n = r^n \operatorname{cis}(n\theta)$
12.28.3
The Power-Reducing Formulas for Sine and Cosine
Rewrite a squared sine or cosine using the cosine of the double angle.
e.g. $\cos^2 15^\circ = \tfrac{1 + \cos 30^\circ}{2} = \tfrac{2 + \sqrt3}{4}$
$\displaystyle \sin^2\theta = \frac{1 - \cos 2\theta}{2}$
$\displaystyle \cos^2\theta = \frac{1 + \cos 2\theta}{2}$
12.28.4
Euler's Formula
Links the exponential function to cosine and sine, with $\theta$ in radians. It gives the exponential form $z = re^{i\theta}$ .
e.g. $2e^{i\pi/3} = 2\left(\tfrac12 + \tfrac{\sqrt3}{2}i\right) = 1 + \sqrt3\,i$
$\displaystyle e^{i\theta} = \cos\theta + i\sin\theta$
$\displaystyle z = re^{i\theta}$
$\displaystyle e^{i\pi} = -1$
12.28.5
The Roots of Unity
The $n$ th roots of unity are the $n$ solutions of $z^n = 1$ . They all have modulus $1$ and are spaced evenly around the unit circle, $\tfrac{2\pi}{n}$ apart, starting at $1$ .
e.g. Cube roots of unity: $1$ , $-\tfrac12 + \tfrac{\sqrt3}{2}i$ , $-\tfrac12 - \tfrac{\sqrt3}{2}i$
$\displaystyle z^n = 1$
$\displaystyle z_k = \operatorname{cis}\frac{2\pi k}{n}$
$\displaystyle k = 0, 1, \dots, n - 1$
12.28.6
Properties of Roots of Unity
With $\omega = \operatorname{cis}\tfrac{2\pi}{n}$ , the $n$ th roots are $1, \omega, \omega^2, \dots, \omega^{n-1}$ . For $n \ge 2$ they add to $0$ . Powers of $\omega$ repeat every $n$ steps, and the conjugate of a root of unity is also a root.
e.g. Cube roots: $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$ , so $\omega^{10} = \omega$
$\displaystyle 1 + \omega + \dots + \omega^{n-1} = 0$
$\displaystyle \omega^n = 1$
$\displaystyle \omega^{k + n} = \omega^k$
12.28.7
Square Roots of Complex Numbers
Every nonzero complex number has two square roots, which are opposites. In polar form, take $\sqrt r$ and halve the argument, then add $\pi$ for the second root. Or set $(a + bi)^2 = z$ and match real and imaginary parts.
e.g. Square roots of $2i$ : $r = 2$ , $\theta = \tfrac{\pi}{2}$ , so the roots are $\pm\sqrt2\operatorname{cis}\tfrac{\pi}{4} = \pm(1 + i)$
$\displaystyle \sqrt r \operatorname{cis}\frac{\theta}{2}$
$\displaystyle \sqrt r \operatorname{cis}\left(\frac{\theta}{2} + \pi\right)$
12.28.8
Higher Roots of Complex Numbers
A nonzero complex number has exactly $n$ distinct $n$ th roots. They share modulus $\sqrt[n]{r}$ and are spaced $\tfrac{2\pi}{n}$ apart around a circle.
e.g. Cube roots of $8i$ ( $r = 8$ , $\theta = \tfrac{\pi}{2}$ ): $2\operatorname{cis}\tfrac{\pi}{6}$ , $2\operatorname{cis}\tfrac{5\pi}{6}$ , $2\operatorname{cis}\tfrac{3\pi}{2}$ , which are $\sqrt3 + i$ , $-\sqrt3 + i$ , $-2i$
$\displaystyle z_k = \sqrt[n]{r}\,\operatorname{cis}\frac{\theta + 2\pi k}{n}$
$\displaystyle k = 0, 1, \dots, n - 1$

12.29 The Fundamental Theorem of Algebra

12.29.1
The Fundamental Theorem of Algebra for Quadratic Equations
Every quadratic has exactly 2 complex roots, counting a repeated root twice. With real coefficients, a negative discriminant gives a pair of complex conjugate roots.
e.g. $x^2 + 2x + 5 = 0$ : $x = \tfrac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i$
$\displaystyle ax^2 + bx + c = a(x - r_1)(x - r_2)$
$\displaystyle b^2 - 4ac < 0 \Rightarrow p \pm qi$
12.29.2
The Fundamental Theorem of Algebra with Cubic Equations
Every cubic has exactly 3 complex roots, counting multiplicity. With real coefficients, nonreal roots come in conjugate pairs, so a real cubic has either 3 real roots or 1 real root and 1 conjugate pair.
e.g. $x^3 - x^2 + 4x - 4 = (x - 1)(x^2 + 4)$ : roots $1$ , $2i$ , $-2i$
$\displaystyle a(x - r_1)(x - r_2)(x - r_3)$
12.29.3
Solving Cubic Equations With Complex Roots
Find one root (test rational candidates, or use a given root). Divide it out with synthetic division, then solve the leftover quadratic. With real coefficients, if $a + bi$ is a root, so is $a - bi$ .
e.g. $x^3 - 3x^2 + 4x - 2 = 0$ : $x = 1$ works, the quotient is $x^2 - 2x + 2$ , so $x = 1 \pm i$
$\displaystyle p(r) = 0 \iff (x - r) \text{ is a factor}$
12.29.4
The Fundamental Theorem of Algebra with Quartic Equations
Every quartic has exactly 4 complex roots, counting multiplicity. With real coefficients, nonreal roots come in conjugate pairs, so there are 4, 2, or 0 real roots.
e.g. $x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)$ : roots $\pm 1$ , $\pm i$
$\displaystyle a(x - r_1)(x - r_2)$
$\displaystyle \cdot (x - r_3)(x - r_4)$
12.29.5
Solving Quartic Equations With Complex Roots
If $a + bi$ is a root, multiply its factor by its conjugate's factor to get a real quadratic factor, then divide it out. Quartics with only $x^4$ , $x^2$ , and constant terms factor like quadratics in $x^2$ .
e.g. $x^4 - 5x^2 - 36 = (x^2 - 9)(x^2 + 4) = 0$ , so $x = \pm 3$ , $\pm 2i$
$\displaystyle (x - (a + bi))(x - (a - bi))$
$\displaystyle = x^2 - 2ax + (a^2 + b^2)$

Unit 13 30 topics

Vectors

Introduction to Vectors, Vectors in 2D Cartesian Coordinates, Vectors in 3D Cartesian Coordinates, and more.

13.30 Introduction to Vectors

To add vectors, put the tail of v at the head of u; the sum runs from the first tail to the last head. It is also the diagonal of the parallelogram the two vectors make.

13.30.1
Introduction to Vectors
A vector has both magnitude and direction. A scalar has only size. A vector is drawn as an arrow from its tail to its head, and two vectors are equal when they have the same magnitude and direction, wherever they sit.
e.g. A speed of 50 mph is a scalar; a velocity of 50 mph due north is a vector.
$\displaystyle \overrightarrow{BA} = -\overrightarrow{AB}$
13.30.2
The Triangle Law for the Addition of Two Vectors
Place the tail of the second vector at the head of the first. The sum runs from the tail of the first to the head of the second. Order does not matter: $\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}$ .
e.g. Walk 3 km east, then 4 km north: you end $\sqrt{3^2 + 4^2} = 5$ km from the start.
$\displaystyle \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$
13.30.3
The Magnitude of a Vector
The magnitude is the length of the vector. It is never negative, and only the zero vector has magnitude $0$ . Scaling by $k$ scales the length by $\lvert k \rvert$ .
e.g. $\lvert \mathbf{v} \rvert = 5 \Rightarrow \lvert -3\mathbf{v} \rvert = 15$
$\displaystyle \lvert \overrightarrow{AB} \rvert = AB$
$\displaystyle \lvert k\mathbf{v} \rvert = \lvert k \rvert\,\lvert \mathbf{v} \rvert$
13.30.4
Problem Solving Using Vector Diagrams
Sketch the vectors head to tail to form a triangle. Use the Pythagorean theorem for a right angle, and the law of cosines or law of sines otherwise.
e.g. A plane heads north at 120 mph with a 50 mph wind from the west: ground speed $\sqrt{120^2 + 50^2} = 130$ mph.
$\displaystyle c^2 = a^2 + b^2 - 2ab\cos C$
$\displaystyle \frac{\sin A}{a} = \frac{\sin B}{b}$
13.30.5
Parallel Vectors
Two nonzero vectors are parallel exactly when one is a scalar multiple of the other. If $k > 0$ they point the same way; if $k < 0$ they point opposite ways.
e.g. $2\mathbf{a} - 4\mathbf{b} = 2(\mathbf{a} - 2\mathbf{b})$ , so it is parallel to $\mathbf{a} - 2\mathbf{b}$ .
$\displaystyle \mathbf{u} = k\mathbf{v}, \quad k \ne 0$
13.30.6
Unit Vectors
A unit vector has magnitude $1$ . Divide any nonzero vector by its magnitude to get the unit vector in its direction. $\mathbf{i}$ and $\mathbf{j}$ are the unit vectors along the $x$ - and $y$ -axes.
e.g. If $\lvert \mathbf{v} \rvert = 4$ , the unit vector is $\tfrac14\mathbf{v}$ , and a vector of length 10 in that direction is $\tfrac{10}{4}\mathbf{v} = 2.5\mathbf{v}$ .
$\displaystyle \hat{\mathbf{v}} = \frac{\mathbf{v}}{\lvert \mathbf{v} \rvert}$
13.30.7
Linear Combinations of Vectors and Their Properties
A linear combination is $a\mathbf{u} + b\mathbf{v}$ for scalars $a$ and $b$ . Vectors follow the usual commutative, associative, and distributive rules. If $\mathbf{u}$ and $\mathbf{v}$ are not parallel and $a\mathbf{u} + b\mathbf{v} = c\mathbf{u} + d\mathbf{v}$ , then $a = c$ and $b = d$ .
e.g. $2(\mathbf{a} + 3\mathbf{b}) - (\mathbf{a} - \mathbf{b}) = \mathbf{a} + 7\mathbf{b}$
$\displaystyle \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$
$\displaystyle k(\mathbf{u} + \mathbf{v}) = k\mathbf{u} + k\mathbf{v}$
$\displaystyle (a + b)\mathbf{v} = a\mathbf{v} + b\mathbf{v}$
13.30.8
Describing the Position Vector of a Point Using Known Vectors
The position vector of a point runs from the origin $O$ to that point. To reach any point, follow a path of known vectors and add them.
e.g. $P$ is $\tfrac13$ of the way from $A$ to $B$ : $\overrightarrow{OP} = \mathbf{a} + \tfrac13(\mathbf{b} - \mathbf{a}) = \tfrac23\mathbf{a} + \tfrac13\mathbf{b}$
$\displaystyle \overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
$\displaystyle \overrightarrow{OM} = \tfrac12(\mathbf{a} + \mathbf{b}) \;\text{(midpoint)}$

13.31 Vectors in 2D Cartesian Coordinates

13.31.1
Two-Dimensional Vectors Expressed in Component Form
A 2D vector is written by its horizontal and vertical changes. For the vector from $A$ to $B$ , subtract the coordinates of the start from the end.
e.g. $A(1, 2)$ , $B(4, -2)$ : $\overrightarrow{AB} = \langle 3, -4 \rangle = 3\mathbf{i} - 4\mathbf{j}$
$\displaystyle \mathbf{v} = \langle a, b \rangle = a\mathbf{i} + b\mathbf{j}$
$\displaystyle \overrightarrow{AB} = \langle x_2 - x_1,\, y_2 - y_1 \rangle$
13.31.2
Addition and Scalar Multiplication of Cartesian Vectors in 2D
Add or subtract matching components. A scalar multiplies every component.
e.g. $2\langle 1, -3 \rangle + \langle 4, 5 \rangle = \langle 6, -1 \rangle$
$\displaystyle \langle a, b \rangle + \langle c, d \rangle$
$\displaystyle = \langle a + c,\, b + d \rangle$
$\displaystyle k\langle a, b \rangle = \langle ka, kb \rangle$
13.31.3
Calculating the Magnitude of Cartesian Vectors in 2D
Use the Pythagorean theorem on the components.
e.g. $\lvert \langle -5, 12 \rangle \rvert = \sqrt{25 + 144} = 13$
$\displaystyle \lvert \langle a, b \rangle \rvert = \sqrt{a^2 + b^2}$
13.31.4
Calculating the Direction of Cartesian Vectors in 2D
The direction is the angle $\theta$ measured counterclockwise from the positive $x$ -axis. Find the reference angle with inverse tangent, then place it in the vector's quadrant: add $180^\circ$ if $a < 0$ , and add $360^\circ$ if the vector is in Quadrant IV.
e.g. $\langle -3, 3 \rangle$ : $\tan^{-1}(-1) = -45^\circ$ , but the vector is in Quadrant II, so $\theta = 135^\circ$
$\displaystyle \tan\theta = \frac{b}{a}$
13.31.5
Calculating the Components of Cartesian Vectors in 2D
Given magnitude and direction angle $\theta$ , the horizontal component uses cosine and the vertical component uses sine.
e.g. Magnitude 10 at $30^\circ$ : $\langle 10\cos 30^\circ, 10\sin 30^\circ \rangle = \langle 5\sqrt3, 5 \rangle$
$\displaystyle \mathbf{v} = \langle \lvert\mathbf{v}\rvert\cos\theta,\, \lvert\mathbf{v}\rvert\sin\theta \rangle$
13.31.6
Velocity and Acceleration for Plane Motion
Velocity and acceleration are vectors; work with each component separately. With constant acceleration $\mathbf{a}$ and initial velocity $\mathbf{u}$ , the velocity at time $t$ is $\mathbf{u} + \mathbf{a}t$ . Speed is the magnitude of velocity.
e.g. $\mathbf{u} = \langle 2, 1 \rangle$ m/s, $\mathbf{a} = \langle 1, -2 \rangle$ m/s $^2$ , $t = 3$ : $\mathbf{v} = \langle 5, -5 \rangle$ , speed $5\sqrt2$ m/s
$\displaystyle \mathbf{v} = \mathbf{u} + \mathbf{a}t$
$\displaystyle \mathbf{a} = \frac{\mathbf{v} - \mathbf{u}}{t}$
$\displaystyle \text{speed} = \lvert \mathbf{v} \rvert$
13.31.7
Calculating Displacement for Plane Motion
Displacement is the change in position. With constant velocity it is $\mathbf{v}t$ . With constant acceleration use the formula below, then add the starting position to get the new position.
e.g. $\mathbf{u} = \langle 4, 0 \rangle$ , $\mathbf{a} = \langle 0, 2 \rangle$ , $t = 3$ : $\mathbf{s} = \langle 12, 0 \rangle + \langle 0, 9 \rangle = \langle 12, 9 \rangle$ , a distance of 15
$\displaystyle \mathbf{s} = \mathbf{u}t + \tfrac12\mathbf{a}t^2$
$\displaystyle \mathbf{r} = \mathbf{r}_0 + \mathbf{s}$

13.32 Vectors in 3D Cartesian Coordinates

13.32.1
Three-Dimensional Vectors in Component Form
A 3D vector has $x$ -, $y$ -, and $z$ -components. $\mathbf{k}$ is the unit vector along the $z$ -axis. Subtract start coordinates from end coordinates.
e.g. $A(1, 0, 2)$ , $B(3, -1, 5)$ : $\overrightarrow{AB} = \langle 2, -1, 3 \rangle$
$\displaystyle \langle a, b, c \rangle = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$
$\displaystyle \overrightarrow{AB} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle$
13.32.2
Addition and Scalar Multiplication of Cartesian Vectors in 3D
Same as 2D: add matching components, and a scalar multiplies every component.
e.g. $\langle 1, 2, -1 \rangle - 2\langle 0, 3, 1 \rangle = \langle 1, -4, -3 \rangle$
$\displaystyle \langle a, b, c \rangle + \langle d, e, f \rangle$
$\displaystyle = \langle a + d,\, b + e,\, c + f \rangle$
$\displaystyle k\langle a, b, c \rangle = \langle ka, kb, kc \rangle$
13.32.3
Calculating the Magnitude of Cartesian Vectors in 3D
Square all three components, add, and take the square root.
e.g. $\lvert \langle 2, 3, 6 \rangle \rvert = \sqrt{4 + 9 + 36} = 7$
$\displaystyle \lvert \langle a, b, c \rangle \rvert = \sqrt{a^2 + b^2 + c^2}$

13.33 The Dot Product

The dot product is the length of v times the length of u’s shadow on v. When the vectors are perpendicular the shadow vanishes, so u · v = 0.

13.33.1
Calculating the Dot Product Using Angle and Magnitude
The dot product is a scalar. Use the angle $\theta$ between the vectors placed tail to tail, with $0^\circ \le \theta \le 180^\circ$ . Perpendicular vectors have dot product $0$ .
e.g. $\lvert\mathbf{u}\rvert = 4$ , $\lvert\mathbf{v}\rvert = 5$ , $\theta = 60^\circ$ : $\mathbf{u}\cdot\mathbf{v} = 20 \cdot \tfrac12 = 10$
$\displaystyle \mathbf{u} \cdot \mathbf{v} = \lvert\mathbf{u}\rvert\,\lvert\mathbf{v}\rvert\cos\theta$
13.33.2
Calculating the Dot Product Using Components
Multiply matching components and add the products. The result is a number, not a vector.
e.g. $\langle 2, -1, 3 \rangle \cdot \langle 4, 5, 1 \rangle = 8 - 5 + 3 = 6$
$\displaystyle \langle a_1, a_2 \rangle \cdot \langle b_1, b_2 \rangle$
$\displaystyle = a_1 b_1 + a_2 b_2$
$\displaystyle \text{3D: add } a_3 b_3$
13.33.3
The Angle Between Two Vectors
Solve the dot product formula for $\cos\theta$ . A dot product of $0$ means perpendicular, positive means acute, and negative means obtuse.
e.g. $\langle 1, 0 \rangle$ and $\langle 1, 1 \rangle$ : $\cos\theta = \tfrac{1}{1 \cdot \sqrt2}$ , so $\theta = 45^\circ$
$\displaystyle \cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{\lvert\mathbf{u}\rvert\,\lvert\mathbf{v}\rvert}$
13.33.4
Calculating a Scalar Projection
The scalar projection of $\mathbf{u}$ onto $\mathbf{v}$ is the signed length of the shadow of $\mathbf{u}$ along $\mathbf{v}$ . It is negative when the angle is obtuse.
e.g. $\mathbf{u} = \langle 2, 1 \rangle$ onto $\mathbf{v} = \langle 3, 4 \rangle$ : $\tfrac{6 + 4}{5} = 2$
$\displaystyle \text{comp}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u}\cdot\mathbf{v}}{\lvert\mathbf{v}\rvert}$
$\displaystyle = \lvert\mathbf{u}\rvert\cos\theta$
13.33.5
Calculating a Vector Projection
The vector projection of $\mathbf{u}$ onto $\mathbf{v}$ is the vector along $\mathbf{v}$ whose length is the scalar projection. Note the squared magnitude in the denominator.
e.g. $\mathbf{u} = \langle 2, 1 \rangle$ onto $\mathbf{v} = \langle 3, 4 \rangle$ : $\tfrac{10}{25}\langle 3, 4 \rangle = \left\langle \tfrac65, \tfrac85 \right\rangle$
$\displaystyle \text{proj}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u}\cdot\mathbf{v}}{\lvert\mathbf{v}\rvert^2}\,\mathbf{v}$

13.34 The Cross Product

The cross product of two 3D vectors is a new vector standing straight up from the plane they span. Its length equals the area of their parallelogram, and the right-hand rule gives its direction.

13.34.1
The Cross Product of Two Vectors
Defined only for vectors in 3D. $\mathbf u \times \mathbf v$ is a vector perpendicular to both $\mathbf u$ and $\mathbf v$ , pointing the way the right-hand rule gives. Its length is $\lVert \mathbf u \rVert \lVert \mathbf v \rVert \sin\theta$ .
e.g. $\langle 1,0,0\rangle \times \langle 0,1,0\rangle = \langle 0,0,1\rangle$ , so $\mathbf i \times \mathbf j = \mathbf k$
$\displaystyle \mathbf u \times \mathbf v = \langle u_2v_3 - u_3v_2,$
$\displaystyle \quad u_3v_1 - u_1v_3,\; u_1v_2 - u_2v_1 \rangle$
$\displaystyle \lVert \mathbf u \times \mathbf v \rVert = \lVert \mathbf u \rVert \lVert \mathbf v \rVert \sin\theta$
13.34.2
Properties of the Cross Product
Swapping the order flips the sign. It distributes over addition and scalars pull out. $\mathbf u \times \mathbf v = \mathbf 0$ exactly when the vectors are parallel, so $\mathbf u \times \mathbf u = \mathbf 0$ . It is not associative.
e.g. $\mathbf i \times \mathbf j = \mathbf k$ , so $\mathbf j \times \mathbf i = -\mathbf k$
$\displaystyle \mathbf v \times \mathbf u = -(\mathbf u \times \mathbf v)$
$\displaystyle \mathbf u \times (\mathbf v + \mathbf w) = \mathbf u \times \mathbf v + \mathbf u \times \mathbf w$
$\displaystyle (c\mathbf u) \times \mathbf v = c(\mathbf u \times \mathbf v)$
$\displaystyle (\mathbf u \times \mathbf v) \cdot \mathbf u = 0$
13.34.3
Calculating the Cross Product Using Determinants
Write a $3 \times 3$ determinant with top row $\mathbf i, \mathbf j, \mathbf k$ , middle row the components of $\mathbf u$ , and bottom row the components of $\mathbf v$ . Expand along the top row: $\mathbf u \times \mathbf v$ is the sum of the three terms below. Keep the minus sign on $\mathbf j$ .
e.g. $\langle 1,2,3\rangle \times \langle 4,5,6\rangle = \mathbf i(12-15) - \mathbf j(6-12) + \mathbf k(5-8) = \langle -3, 6, -3\rangle$
$\displaystyle \mathbf i \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix}$
$\displaystyle - \mathbf j \begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix}$
$\displaystyle + \mathbf k \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix}$
13.34.4
Finding Areas Using the Cross Product
The length of $\mathbf u \times \mathbf v$ is the area of the parallelogram with sides $\mathbf u$ and $\mathbf v$ . A triangle is half of that parallelogram. For a triangle $PQR$ , use two edge vectors from the same vertex.
e.g. $P(1,0,0)$ , $Q(2,2,0)$ , $R(4,1,0)$ : $\langle 1,2,0\rangle \times \langle 3,1,0\rangle = \langle 0,0,-5\rangle$ , so area $= \tfrac52$
$\displaystyle A_{\text{par}} = \lVert \mathbf u \times \mathbf v \rVert$
$\displaystyle A_{\text{tri}} = \frac12 \lVert \overrightarrow{PQ} \times \overrightarrow{PR} \rVert$
13.34.5
The Scalar Triple Product
$\mathbf u \cdot (\mathbf v \times \mathbf w)$ is a number, not a vector. It equals the determinant of the $3 \times 3$ matrix whose rows are $\mathbf u$ , $\mathbf v$ , $\mathbf w$ . Cycling the order keeps the value; swapping two vectors flips the sign. It is $0$ exactly when the three vectors are coplanar.
e.g. $\mathbf u = \langle 1,2,0\rangle$ , $\mathbf v = \langle 0,1,3\rangle$ , $\mathbf w = \langle 2,0,1\rangle$ : $\mathbf v \times \mathbf w = \langle 1,6,-2\rangle$ , so $\mathbf u \cdot (\mathbf v \times \mathbf w) = 1 + 12 + 0 = 13$
$\displaystyle \mathbf u \cdot (\mathbf v \times \mathbf w)$
$\displaystyle = \mathbf v \cdot (\mathbf w \times \mathbf u)$
$\displaystyle = \mathbf w \cdot (\mathbf u \times \mathbf v)$
13.34.6
Volumes of Parallelepipeds
The volume of the parallelepiped with edges $\mathbf u$ , $\mathbf v$ , $\mathbf w$ from one corner is the absolute value of the scalar triple product. A volume of $0$ means the edges are coplanar.
e.g. Edges $\langle 2,0,0\rangle$ , $\langle 1,3,0\rangle$ , $\langle 0,1,4\rangle$ : $\langle 1,3,0\rangle \times \langle 0,1,4\rangle = \langle 12,-4,1\rangle$ , so $V = \lvert 24 + 0 + 0 \rvert = 24$
$\displaystyle V = \lvert \mathbf u \cdot (\mathbf v \times \mathbf w) \rvert$
13.34.7
Finding Volumes of Tetrahedrons and Pyramids Using Vector Products
Take three edge vectors from one vertex. A tetrahedron is $\tfrac16$ of the parallelepiped. A pyramid whose base is the parallelogram on $\mathbf u$ , $\mathbf v$ , with $\mathbf w$ running from that corner to the apex, is $\tfrac13$ of it.
e.g. Vertices $(0,0,0)$ , $(2,0,0)$ , $(0,3,0)$ , $(0,0,4)$ : triple product $= 24$ , so $V = \tfrac{24}{6} = 4$
$\displaystyle V_{\text{tet}} = \frac16 \lvert \mathbf u \cdot (\mathbf v \times \mathbf w) \rvert$
$\displaystyle V_{\text{pyr}} = \frac13 \lvert \mathbf u \cdot (\mathbf v \times \mathbf w) \rvert$

Unit 14 37 topics

Matrices

Introduction to Matrices, Matrix Multiplication, Determinants, and more.

14.35 Introduction to Matrices

14.35.1
Introduction to Matrices
A matrix is a rectangular array of numbers called entries. Its size is (rows) $\times$ (columns), always rows first.
e.g. $\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ is a $2 \times 3$ matrix
$\displaystyle m \times n = \text{rows} \times \text{columns}$
14.35.2
Index Notation for Matrices
$a_{ij}$ is the entry of matrix $A$ in row $i$ and column $j$ , row first.
e.g. For $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ : $a_{12} = 2$ and $a_{23} = 6$
$\displaystyle a_{ij} = \text{entry in row } i,\ \text{column } j$
14.35.3
Adding and Subtracting Matrices
Only matrices of the same size can be added or subtracted. Combine the entries in matching positions.
e.g. $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} + \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 6 & 1 \\ 3 & 6 \end{pmatrix}$
$\displaystyle (A \pm B)_{ij} = a_{ij} \pm b_{ij}$
14.35.4
Properties of Matrix Addition
For matrices of the same size, addition is commutative and associative. The zero matrix $O$ is the additive identity, and $-A$ undoes $A$ .
e.g. $A - B = A + (-B)$
$\displaystyle A + B = B + A$
$\displaystyle (A + B) + C = A + (B + C)$
$\displaystyle A + O = A$
$\displaystyle A + (-A) = O$
14.35.5
Scalar Multiplication of Matrices
Multiply every entry by the scalar. Scalars distribute over matrix addition.
e.g. $3\begin{pmatrix} 1 & -2 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 3 & -6 \\ 0 & 12 \end{pmatrix}$
$\displaystyle (kA)_{ij} = k\,a_{ij}$
$\displaystyle k(A + B) = kA + kB$
14.35.6
Zero, Square, Diagonal and Identity Matrices
A zero matrix $O$ has every entry $0$ . A square matrix is $n \times n$ . A diagonal matrix is square with $0$ everywhere off the main diagonal (top-left to bottom-right). The identity $I_n$ is diagonal with every diagonal entry $1$ .
e.g. $\begin{pmatrix} 2 & 0 \\ 0 & -5 \end{pmatrix}$ is diagonal but not the identity
$\displaystyle I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
14.35.7
The Transpose of a Matrix
Rows become columns: row $i$ of $A$ becomes column $i$ of $A^T$ . An $m \times n$ matrix becomes $n \times m$ .
e.g. $\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}^T = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix}$
$\displaystyle (A^T)_{ij} = a_{ji}$
$\displaystyle (A^T)^T = A$
$\displaystyle (A + B)^T = A^T + B^T$

14.36 Matrix Multiplication

14.36.1
Multiplying a Matrix by a Column Vector
Each entry of the result is a row of the matrix dotted with the vector. The number of columns must equal the number of entries in the vector.
e.g. $\begin{pmatrix} 2 & 1 \\ 3 & -1 \end{pmatrix}\begin{pmatrix} 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 8 + 5 \\ 12 - 5 \end{pmatrix} = \begin{pmatrix} 13 \\ 7 \end{pmatrix}$
$\displaystyle \begin{pmatrix} a & b \\ c & d \end{pmatrix} \binom{x}{y}$
$\displaystyle = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}$
14.36.2
Multiplying Square Matrices
The entry in row $i$ , column $j$ of $AB$ is row $i$ of $A$ dotted with column $j$ of $B$ .
e.g. $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 2 + 2 & 0 + 6 \\ 6 + 4 & 0 + 12 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 10 & 12 \end{pmatrix}$
$\displaystyle (AB)_{ij} = (\text{row } i \text{ of } A)$
$\displaystyle \quad \cdot\, (\text{column } j \text{ of } B)$
14.36.3
Conformability for Matrix Multiplication
$AB$ is defined only when the number of columns of $A$ equals the number of rows of $B$ . The result takes the outer dimensions.
e.g. $(2 \times 3)(3 \times 4)$ gives $2 \times 4$ ; $(2 \times 3)(2 \times 3)$ is undefined
$\displaystyle (m \times n)(n \times p) \to m \times p$
14.36.4
Multiplying Matrices
Check conformability first. Then each entry of $AB$ is a row of $A$ dotted with a column of $B$ .
e.g. $\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 1 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 2 & 13 \end{pmatrix}$
$\displaystyle (AB)_{ij} = a_{i1}b_{1j} + \cdots + a_{in}b_{nj}$
14.36.5
Powers of Matrices
$A^n$ means $A$ multiplied by itself $n$ times, so only square matrices have powers. $A^2$ is $AA$ , not the entries squared.
e.g. $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$
$\displaystyle A^n = \underbrace{AA \cdots A}_{n \text{ factors}}$
14.36.6
Multiplying a Matrix by the Identity Matrix
Multiplying by an identity matrix of the right size leaves a matrix unchanged. For an $m \times n$ matrix $A$ , use $I_m$ on the left and $I_n$ on the right.
e.g. $\begin{pmatrix} 3 & 1 \\ 2 & 5 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 2 & 5 \end{pmatrix}$
$\displaystyle AI = IA = A$
$\displaystyle I_m A = A I_n = A$
14.36.7
Properties of Matrix Multiplication
Matrix multiplication is associative and distributive, and scalars can move anywhere. It is not commutative: usually $AB \ne BA$ , so keep the order.
e.g. $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ : $AB = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ but $BA = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$
$\displaystyle (AB)C = A(BC)$
$\displaystyle A(B + C) = AB + AC$
$\displaystyle k(AB) = (kA)B = A(kB)$
$\displaystyle AB \ne BA \;\text{in general}$
14.36.8
Representing 2x2 Systems of Equations Using a Matrix Product
Put the coefficients in a matrix $A$ , the variables in a column vector $\mathbf x$ , and the constants in a column vector $\mathbf b$ . The system becomes $A\mathbf x = \mathbf b$ .
e.g. $2x + 3y = 7$ , $x - y = 1$ becomes $\begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}$
$\displaystyle \begin{cases} ax + by = e \\ cx + dy = f \end{cases}$
$\displaystyle \Leftrightarrow\; \begin{pmatrix} a & b \\ c & d \end{pmatrix} \binom{x}{y} = \binom{e}{f}$
14.36.9
Representing 3x3 Systems of Equations Using a Matrix Product
Same idea with a $3 \times 3$ coefficient matrix and $\mathbf x = (x, y, z)$ as a column. Line the variables up in the same order in every equation, and write $0$ for a missing variable.
e.g. $x + 2y - z = 3$ , $2x + y = 1$ , $y + 3z = 4$ becomes $\begin{pmatrix} 1 & 2 & -1 \\ 2 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}$
$\displaystyle A\mathbf x = \mathbf b$

14.37 Determinants

14.37.1
The Determinant of a 2x2 Matrix
Main diagonal product minus the other diagonal product. It is written $\det A$ or with vertical bars. Only square matrices have determinants.
e.g. $\det\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} = 12 - 2 = 10$
$\displaystyle \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$
14.37.2
The Geometric Interpretation of the 2x2 Determinant
$\lvert \det A \rvert$ is the area of the parallelogram formed by the two column vectors of $A$ . A negative determinant means the vectors are in clockwise order. A determinant of $0$ means the columns are parallel, so the area is $0$ .
e.g. Columns $(3, 0)$ and $(1, 2)$ : $\det\begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} = 6 - 0 = 6$ , so area $6$
$\displaystyle \text{Area} = \lvert ad - bc \rvert$
14.37.3
The Minors of a 3x3 Matrix
The minor $M_{ij}$ is the determinant of the $2 \times 2$ matrix left after deleting row $i$ and column $j$ .
e.g. For $\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$ : $M_{11} = \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = 45 - 48 = -3$
$\displaystyle M_{ij} = \det(\text{delete row } i,\ \text{column } j)$
14.37.4
The Determinant of a 3x3 Matrix
Expand along the top row: each entry times its minor, with signs $+,\,-,\,+$ . For rows $(a, b, c)$ , $(d, e, f)$ , $(g, h, i)$ :
e.g. $\det\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{pmatrix} = 1(50 - 48) - 2(40 - 42) + 3(32 - 35) = 2 + 4 - 9 = -3$
$\displaystyle \det A = aM_{11} - bM_{12} + cM_{13}$
$\displaystyle = a(ei - fh) - b(di - fg)$
$\displaystyle \quad + c(dh - eg)$

14.38 The Inverse of a Matrix

14.38.1
Introduction to the Inverse of a Matrix
The inverse $A^{-1}$ undoes $A$ : their product either way is the identity. Only square matrices can have inverses, and $A^{-1}$ exists exactly when $\det A \ne 0$ . A matrix with $\det A = 0$ is singular.
e.g. $\begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}^{-1} = \begin{pmatrix} \tfrac12 & 0 \\ 0 & \tfrac14 \end{pmatrix}$
$\displaystyle AA^{-1} = A^{-1}A = I$
14.38.2
Inverses of 2x2 Matrices
Swap $a$ and $d$ , negate $b$ and $c$ , then divide by the determinant. If $ad - bc = 0$ there is no inverse.
e.g. $\begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}^{-1} = \tfrac{1}{10}\begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix}$ , since $\det = 24 - 14 = 10$
$\displaystyle A^{-1} = \frac{1}{\det A}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$
$\displaystyle \det A = ad - bc \ne 0$
14.38.3
Calculating the Inverse of a 3x3 Matrix Using the Cofactor Method
Find all nine minors. Apply the checkerboard signs ( $+$ in the top-left corner, alternating) to get the cofactors. Transpose the cofactor matrix to get the adjugate, then divide by $\det A$ .
e.g. $A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ : $\det A = 1$ , diagonal cofactors are $1$ , $C_{21} = -2$ , the rest $0$ ; transposing gives $A^{-1} = \begin{pmatrix} 1 & -2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
$\displaystyle C_{ij} = (-1)^{i+j} M_{ij}$
$\displaystyle \operatorname{adj} A = C^T$
$\displaystyle A^{-1} = \frac{1}{\det A}\operatorname{adj} A$
14.38.4
Solving 2x2 Systems of Equations Using Inverse Matrices
Write the system as $A\mathbf x = \mathbf b$ , then multiply both sides on the left by $A^{-1}$ . This works only when $\det A \ne 0$ .
e.g. $2x + y = 5$ , $x + y = 3$ : $A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$ , so $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 - 3 \\ -5 + 6 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$
$\displaystyle A\mathbf x = \mathbf b$
$\displaystyle \Rightarrow\; \mathbf x = A^{-1}\mathbf b$
14.38.5
Solving Systems of Equations Using Inverse Matrices
Any square system $A\mathbf x = \mathbf b$ with $\det A \ne 0$ has the unique solution $\mathbf x = A^{-1}\mathbf b$ . Order matters: $A^{-1}$ goes on the left of $\mathbf b$ . If $\det A = 0$ there is no unique solution.
e.g. $x + 2y = 5$ , $y = 2$ , $z = 3$ : with $A^{-1}$ from the cofactor example, $\mathbf x = \begin{pmatrix} 5 - 4 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$
$\displaystyle \mathbf x = A^{-1}\mathbf b$

14.39 Linear Transformations

A matrix moves î and ĵ to its columns, and every other point follows along. This one shears and stretches the unit square into a parallelogram with twice the area, which is its determinant.

14.39.1
Introduction to Linear Transformations
A linear transformation $T$ maps vectors to vectors and preserves addition and scalar multiples. It always sends the origin to the origin and lines to lines (or to a single point). Every linear transformation of the plane is multiplication by a $2 \times 2$ matrix.
e.g. $T(x, y) = (2x, x + y)$ is linear. $T(x, y) = (x + 1, y)$ is not, since $T(0, 0) = (1, 0)$ .
$\displaystyle T(\mathbf u + \mathbf v) = T(\mathbf u) + T(\mathbf v)$
$\displaystyle T(k\mathbf v) = k\,T(\mathbf v)$
14.39.2
The Standard Matrix of a Linear Transformation
The columns of the matrix are the images of the unit vectors $\mathbf i = (1, 0)$ and $\mathbf j = (0, 1)$ , in that order.
e.g. $T(x, y) = (2x, x + y)$ : $T(\mathbf i) = (2, 1)$ , $T(\mathbf j) = (0, 1)$ , so $A = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}$
$\displaystyle A = \begin{pmatrix} T(\mathbf i) & T(\mathbf j) \end{pmatrix}$
14.39.3
Linear Transformations of Points and Lines in the Plane
To transform a point, multiply its position vector by the matrix. To transform a line, transform a general point on it (or two points) and find the line through the images.
e.g. $A = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}$ : $(3, -1) \mapsto (6, 2)$ . The line $y = 2x$ : $(t, 2t) \mapsto (2t, 3t)$ , so its image is $y = \tfrac32 x$ .
$\displaystyle \binom{x'}{y'} = A\binom{x}{y}$
14.39.4
Linear Transformations of Objects in the Plane
Transform each vertex and join the images in the same order. Faster: put the vertices as the columns of one matrix and multiply once.
e.g. Unit square under $\begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}$ : $(0,0), (1,0), (1,1), (0,1) \mapsto (0,0), (2,1), (2,2), (0,1)$ , a parallelogram
$\displaystyle A\begin{pmatrix} x_1 & x_2 & \cdots \\ y_1 & y_2 & \cdots \end{pmatrix}$
14.39.5
Dilations and Reflections as Linear Transformations
A dilation about the origin with scale factor $k$ multiplies every coordinate by $k$ . The four standard reflections are in the $x$ -axis, the $y$ -axis, $y = x$ , and $y = -x$ .
e.g. Reflecting $(2, 5)$ in $y = x$ gives $(5, 2)$
$\displaystyle \text{scale } k: \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$
$\displaystyle x\text{-axis}: \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
$\displaystyle y\text{-axis}: \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
$\displaystyle y = x: \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
$\displaystyle y = -x: \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$
14.39.6
Shear and Stretch as Linear Transformations
A stretch parallel to the $x$ -axis with factor $k$ multiplies $x$ by $k$ and keeps the $y$ -axis fixed (likewise for $y$ ). An $x$ -shear keeps the $x$ -axis fixed and slides each point sideways by $k$ times its $y$ -coordinate (likewise for a $y$ -shear).
e.g. The shear $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ sends $(1, 3)$ to $(1 + 6, 3) = (7, 3)$
$\displaystyle x\text{-stretch}: \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$
$\displaystyle y\text{-stretch}: \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$
$\displaystyle x\text{-shear}: \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$
$\displaystyle y\text{-shear}: \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}$
14.39.7
Right-Angle Rotations as Linear Transformations
Rotations about the origin are counterclockwise unless stated. $90^\circ$ sends $(x, y)$ to $(-y, x)$ ; $180^\circ$ sends it to $(-x, -y)$ ; $270^\circ$ ( $90^\circ$ clockwise) sends it to $(y, -x)$ .
e.g. Rotating $(3, 1)$ by $90^\circ$ gives $(-1, 3)$
$\displaystyle 90^\circ: \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
$\displaystyle 180^\circ: \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$
$\displaystyle 270^\circ: \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$
14.39.8
Rotations as Linear Transformations
Rotation about the origin by angle $\theta$ counterclockwise. For a clockwise rotation use $-\theta$ .
e.g. Rotating $(2, 0)$ by $60^\circ$ : $(2\cos 60^\circ, 2\sin 60^\circ) = (1, \sqrt3)$
$\displaystyle R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$
14.39.9
Combining Linear Transformations Using 2x2 Matrices
To apply $A$ first and then $B$ , use the single matrix $BA$ . The first transformation goes on the right, next to the vector.
e.g. Reflect in the $x$ -axis, then rotate $90^\circ$ : $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ , a reflection in $y = x$
$\displaystyle \mathbf x \mapsto B(A\mathbf x) = (BA)\mathbf x$
14.39.10
Inverting Linear Transformations
The inverse transformation has matrix $A^{-1}$ and maps each image back to its original point. It exists only when $\det A \ne 0$ . To undo a combination, undo the steps in reverse order.
e.g. Under $A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ the image is $(5, 3)$ : $A^{-1}\begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$
$\displaystyle \mathbf x = A^{-1}\mathbf x'$
$\displaystyle (BA)^{-1} = A^{-1}B^{-1}$
14.39.11
Area Scale Factors of Linear Transformations
Every area is multiplied by $\lvert \det A \rvert$ . A negative determinant also means the shape is flipped (orientation reversed).
e.g. $A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}$ has $\det A = 6 - 1 = 5$ , so a triangle of area $4$ maps to area $20$
$\displaystyle \text{image area} = \lvert \det A \rvert \times \text{area}$
14.39.12
Singular Linear Transformations in the Plane
A transformation is singular when $\det A = 0$ . It squashes the whole plane onto a line through the origin (or onto the origin itself), areas become $0$ , many points share an image, and there is no inverse.
e.g. $\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ : $\det = 4 - 4 = 0$ , and $(x, y) \mapsto (x + 2y,\, 2x + 4y)$ always lands on the line $y = 2x$
$\displaystyle \det A = ad - bc = 0$

Unit 15 19 topics

Probability and Statistics

Regression, Discrete Random Variables, and The Normal Distribution.

15.40 Regression

A residual is the vertical gap between a data point and the fitted line. The least-squares line is the one that makes the squared gaps add up to as little as possible.

15.40.1
Linear Regression
The least-squares regression line is the line that best fits the data by making the sum of the squared residuals as small as possible. It always passes through $(\bar x, \bar y)$ . Find it with a calculator or Desmos.
e.g. $r = 0.8$ , $s_x = 2$ , $s_y = 5$ , $\bar x = 4$ , $\bar y = 10$ : $b = 0.8 \cdot \tfrac52 = 2$ , $a = 10 - 8 = 2$ , so $\hat y = 2x + 2$
$\displaystyle \hat y = a + bx$
$\displaystyle b = r\,\frac{s_y}{s_x}$
$\displaystyle a = \bar y - b\bar x$
15.40.2
Quadratic Regression
Fits a parabola to data that rise then fall, or fall then rise. Use a calculator or Desmos (type $y_1 \sim ax_1^2 + bx_1 + c$ ). An $R^2$ close to 1 means a good fit.
e.g. Points $(0, 1)$ , $(1, 2)$ , $(2, 5)$ : $c = 1$ , $a + b = 1$ , $4a + 2b = 4$ , so $a = 1$ , $b = 0$ and $y = x^2 + 1$
$\displaystyle \hat y = ax^2 + bx + c$
15.40.3
Semi-Log Scatter Plots
Plot $\log y$ against $x$ . If the points lie close to a straight line, the original data are exponential. The intercept gives $\log a$ and the slope gives $\log b$ .
e.g. Line $\log y = 0.3x + 1$ : $a = 10^1 = 10$ , $b = 10^{0.3} \approx 2$ , so $y \approx 10 \cdot 2^x$
$\displaystyle y = ab^x$
$\displaystyle \Rightarrow\; \log y = \log a + x \log b$
15.40.4
Exponential Regression
Fits $y = ab^x$ to data that grow or shrink by a constant factor. Use a calculator or Desmos, or fit a line to the points $(x, \ln y)$ and convert back. $b > 1$ means growth and $0 < b < 1$ means decay.
e.g. Points $(0, 3)$ , $(1, 6)$ , $(2, 12)$ double each step: $y = 3 \cdot 2^x$
$\displaystyle \hat y = ab^x$
$\displaystyle \ln y = \ln a + x \ln b$

15.41 Discrete Random Variables

A binomial distribution counts successes in n independent trials. With p = 0.5 the bars are symmetric and peak at the mean, np = 3.

15.41.1
Probability Mass Functions of Discrete Random Variables
A probability mass function (PMF) gives $P(X = x)$ for each value a discrete random variable can take. Every probability is between 0 and 1, and all of them add to 1.
e.g. $P(X = 1) = 0.2$ , $P(X = 2) = 0.5$ , $P(X = 3) = k$ : $k = 1 - 0.7 = 0.3$
$\displaystyle 0 \le P(X = x) \le 1$
$\displaystyle \sum P(X = x) = 1$
15.41.2
Cumulative Distribution Functions for Discrete Random Variables
The CDF $F(x)$ is the probability that $X$ is at most $x$ . Add the PMF values up to and including $x$ . It never decreases and ends at 1.
e.g. PMF $0.2, 0.5, 0.3$ on $1, 2, 3$ : $F(2) = 0.2 + 0.5 = 0.7$ , so $P(X > 2) = 0.3$
$\displaystyle F(x) = P(X \le x)$
$\displaystyle P(a < X \le b) = F(b) - F(a)$
15.41.3
Expected Values of Discrete Random Variables
Multiply each value by its probability and add. The expected value is the long-run average of $X$ , and it need not be a possible value.
e.g. PMF $0.2, 0.5, 0.3$ on $1, 2, 3$ : $E(X) = 0.2 + 1 + 0.9 = 2.1$
$\displaystyle E(X) = \sum x\,P(X = x)$
15.41.4
The Binomial Distribution
$X \sim B(n, p)$ counts the successes in $n$ independent trials, each with the same success probability $p$ . $k$ can be $0, 1, \ldots, n$ .
e.g. $n = 4$ , $p = \tfrac12$ : $P(X = 2) = 6 \cdot \tfrac{1}{16} = \tfrac38$
$\displaystyle P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
$\displaystyle E(X) = np$
$\displaystyle \text{Var}(X) = np(1-p)$
15.41.5
Modeling With the Binomial Distribution
Use it when there is a fixed number of trials, each trial is a success or failure, trials are independent, and $p$ stays the same. For "at least one," use the complement.
e.g. 3 free throws, $p = 0.8$ : $P(\text{all 3}) = 0.8^3 = 0.512$ ; $P(\text{at least 1}) = 1 - 0.2^3 = 0.992$
$\displaystyle P(X \ge 1) = 1 - P(X = 0)$
$\displaystyle P(X \le k) = \sum_{i=0}^{k} P(X = i)$
15.41.6
The Geometric Distribution
$X \sim \text{Geo}(p)$ is the number of the trial on which the first success happens, with independent trials and constant success probability $p$ . $k$ can be $1, 2, 3, \ldots$
e.g. First 6 on the 3rd roll of a die: $\left(\tfrac56\right)^2 \cdot \tfrac16 = \tfrac{25}{216}$
$\displaystyle P(X = k) = (1-p)^{k-1} p$
$\displaystyle P(X > k) = (1-p)^k$
$\displaystyle E(X) = \frac1p$
15.41.7
Modeling With the Geometric Distribution
Use it when you repeat independent trials with the same $p$ until the first success. "Within $k$ tries" means $X \le k$ , which is easiest by the complement.
e.g. $p = 0.3$ : success within 2 tries is $1 - 0.7^2 = 0.51$ ; expected tries $= \tfrac{1}{0.3} \approx 3.33$
$\displaystyle P(X \le k) = 1 - (1-p)^k$

15.42 The Normal Distribution

In a normal distribution, about 68% of values fall within one standard deviation of the mean, 95% within two, and 99.7% within three. A z-score says how many standard deviations a value sits from the mean.

15.42.1
The Z-Score
A $z$ -score says how many standard deviations a value is from the mean. Positive means above the mean and negative means below.
e.g. $\mu = 70$ , $\sigma = 5$ , $x = 80$ : $z = \tfrac{10}{5} = 2$
$\displaystyle z = \frac{x - \mu}{\sigma}$
15.42.2
The Standard Normal Distribution
$Z \sim N(0, 1)$ is the bell curve with mean 0 and standard deviation 1. The total area under it is 1. $\Phi(z) = P(Z < z)$ is the area to the left of $z$ , from a table or calculator.
e.g. $P(0 < Z < 1) = 0.8413 - 0.5 = 0.3413$
$\displaystyle Z \sim N(0, 1)$
$\displaystyle P(Z < z) = \Phi(z)$
$\displaystyle P(a < Z < b) = \Phi(b) - \Phi(a)$
15.42.3
Symmetry Properties of the Standard Normal Distribution
The curve is symmetric about 0. The area to the left of $-z$ equals the area to the right of $z$ , and each half has area 0.5.
e.g. $\Phi(1) = 0.8413$ : $\Phi(-1) = 0.1587$ and $P(-1 < Z < 1) = 0.6826$
$\displaystyle \Phi(-z) = 1 - \Phi(z)$
$\displaystyle P(Z > z) = 1 - \Phi(z)$
$\displaystyle P(-z < Z < z) = 2\Phi(z) - 1$
15.42.4
The Normal Distribution
$X \sim N(\mu, \sigma^2)$ is a bell curve centered at $\mu$ with standard deviation $\sigma$ . Find probabilities by converting to $z$ -scores. $P(X = a) = 0$ , so $<$ and $\le$ give the same answer.
e.g. $X \sim N(50, 4^2)$ : $P(X < 54) = \Phi(1) \approx 0.8413$
$\displaystyle X \sim N(\mu, \sigma^2)$
$\displaystyle P(X < x) = \Phi\left(\frac{x - \mu}{\sigma}\right)$
15.42.5
Mean and Variance of the Normal Distribution
For $X \sim N(\mu, \sigma^2)$ , the mean is $\mu$ , the variance is $\sigma^2$ , and the standard deviation is $\sigma$ . The mean, median, and mode all sit at the peak. To find an unknown $\mu$ or $\sigma$ , get $z$ from the given probability and solve $x = \mu + z\sigma$ .
e.g. $\sigma = 3$ , $P(X < 26) = 0.9772$ : $z = 2$ , so $26 = \mu + 6$ and $\mu = 20$
$\displaystyle E(X) = \mu$
$\displaystyle \text{Var}(X) = \sigma^2$
$\displaystyle x = \mu + z\sigma$
15.42.6
Percentage Points of the Standard Normal Distribution
Work backward from an area to a $z$ -value: find $z$ with $P(Z < z) = p$ using an inverse normal table or invNorm. For an area to the right, use $1 - p$ on the left.
e.g. $P(Z < z) = 0.975 \Rightarrow z \approx 1.96$ ; $P(Z > z) = 0.05 \Rightarrow z \approx 1.645$
$\displaystyle P(Z < z) = p \;\Rightarrow\; z = \Phi^{-1}(p)$
15.42.7
Modeling With the Normal Distribution
Standardize with $z = \tfrac{x - \mu}{\sigma}$ , then use the standard normal. To turn a $z$ -value back into a data value, use $x = \mu + z\sigma$ .
e.g. Heights $N(64, 3^2)$ in: $P(X > 67) = P(Z > 1) \approx 0.1587$
$\displaystyle z = \frac{x - \mu}{\sigma}$
$\displaystyle x = \mu + z\sigma$
15.42.8
The Empirical Rule for the Normal Distribution
For any normal distribution, about 68%, 95%, and 99.7% of values lie within 1, 2, and 3 standard deviations of the mean.
e.g. $\mu = 100$ , $\sigma = 15$ : about 95% lie between 70 and 130; $P(X > 115) \approx \tfrac{1 - 0.68}{2} = 0.16$
$\displaystyle \mu \pm \sigma: \; 68\%$
$\displaystyle \mu \pm 2\sigma: \; 95\%$
$\displaystyle \mu \pm 3\sigma: \; 99.7\%$

1.1 Finite Geometric Series

The Sum of a Finite Geometric Series

The Sum of the First N Terms of a Geometric Series

Writing Geometric Series in Sigma Notation

Sums of Finite Geometric Series Given in Sigma Notation

Solving Geometric Series Problems Using Exponential Equations and Inequalities

Modeling With Geometric Series

Modeling Financial Problems Using Geometric Series

2.2 Dividing Polynomials

Dividing Polynomials Using Synthetic Division

Dividing Polynomials by Linear Binomials Using Long Division

Dividing Polynomials Using Long Division

Dividing Polynomials by Manipulating Rational Expressions

Closure Properties of Polynomials

2.3 Polynomial Theorems

The Factor Theorem

Determining Polynomial Coefficients Using the Factor Theorem

Factoring Cubic Polynomials Using the Factor Theorem

Factoring Quartic Polynomials Using the Factor Theorem

Multiplicities of the Roots of Polynomials

Finding Multiplicities of the Roots of Quartic Polynomials by Factoring

The Remainder Theorem

The Rational Roots Theorem

2.4 Graphs of Polynomials

Graphing Elementary Cubic Functions

Graphing Cubic Curves Containing Three Distinct Real Roots

Graphing Cubic Curves Containing a Double Root

Graphing Cubic Curves Containing One Distinct Real Root

End Behavior of Polynomials

Graphing General Polynomials

3.5 Rational Equations

Rational Equations With Three Terms

Advanced Rational Equations

Further Advanced Rational Equations

3.6 Reciprocal Functions

Graphing Reciprocal Functions

Graph Transformations of Reciprocal Functions

Combining Graph Transformations of Reciprocal Functions

Domain and Range of Transformed Reciprocal Functions

Inverses of Reciprocal Functions

Finding Intersections of Lines and Reciprocal Functions

3.7 Rational Functions

Roots of Rational Functions

Vertical Asymptotes of Rational Functions

Locating Holes in Rational Functions

Horizontal Asymptotes of Rational Functions

End Behavior of Rational Functions

Infinite Limits of Rational Functions

Infinite Limits of Rational Functions: Advanced Cases

The Domain and Range of a Rational Function

Identifying a Rational Function From a Graph

Identifying a Rational Function From a Graph Containing Holes

Identifying the Graph of a Rational Function

Closure Properties of Rational Expressions

4.8 Radical Expressions

Simplifying Square Root Expressions Using Polynomial Factorization

Rationalizing Denominators of Algebraic Expressions

Rationalizing Denominators With Two Terms

Further Rationalizing Denominators of Algebraic Expressions

4.9 Graphs of Radical Functions

Graphing the Square Root Function

Graph Transformations of Square Root Functions

Graphing the Cube Root Function

Properties of Transformed Square Root Functions

The Domain of a Transformed Radical Function

The Range of a Transformed Radical Function

Roots of Transformed Radical Functions

Inverses of Radical Functions

Finding Intersections of Lines and Radical Functions

5.10 Quadratic Inequalities

Solving Elementary Quadratic Inequalities

Solving Quadratic Inequalities From Graphs

Solving Quadratic Inequalities Using the Graphical Method

Solving Quadratic Inequalities Using the Sign Table Method

Determining Unknown Parameters in Quadratic Equations With Real Solutions

Determining Unknown Parameters in Quadratic Equations With No Real Solutions

Determining Unknown Parameters in Quadratic Equations: Advanced Cases

5.11 Polynomial Inequalities

Inequalities Involving Powers of Binomials

Solving Polynomial Inequalities Using a Graphical Method