SAT Math Prep Reference

Unit 1 77 topics

Algebra

Linear Expressions and Equations in One Variable, Word Problems and Modeling, Two-Variable Linear Equations, and more.

1.1 Linear Expressions and Equations in One Variable

An equation is a balance. Whatever you do to one side, do to the other.

1.1.1
Evaluating Expressions
Substitute each value in parentheses, then follow the order of operations (PEMDAS).
e.g. $3x^2 - 2y$ at $x = -2$ , $y = 5$ : $3(4) - 10 = 2$
1.1.2
Finding Equivalent Equations
Doing the same operation to both sides gives an equivalent equation with the same solutions. You may add, subtract, multiply, or divide by any nonzero number.
e.g. $2x + 6 = 10$ is equivalent to $x + 3 = 5$ (divide by 2)
$\displaystyle a = b \Rightarrow a + c = b + c$
$\displaystyle a = b \Rightarrow ka = kb \;(k \ne 0)$
1.1.3
Solving One-Step Addition Equations
Subtract the added number from both sides.
e.g. $x + 7 = 3 \Rightarrow x = -4$
$\displaystyle x + a = b \Rightarrow x = b - a$
1.1.4
Solving One-Step Multiplication Equations
Divide both sides by the coefficient. If $x$ is divided by a number, multiply both sides by it.
e.g. $-4x = 28 \Rightarrow x = -7$
$\displaystyle ax = b \Rightarrow x = \frac{b}{a} \;(a \ne 0)$
$\displaystyle \frac{x}{a} = b \Rightarrow x = ab$
1.1.5
Solving Multistep Linear Equations
Undo addition and subtraction first, then multiplication and division. Check by substituting back.
e.g. $3x - 5 = 16 \Rightarrow 3x = 21 \Rightarrow x = 7$
1.1.6
Solving Equations by Combining Terms
Combine like terms on each side first. Then move the variable terms to one side and the constants to the other.
e.g. $4x + 3 - x = 2x + 8 \Rightarrow 3x + 3 = 2x + 8 \Rightarrow x = 5$
1.1.7
Solving Equations With Fractional Coefficients
Multiply every term on both sides by the LCD to clear the fractions, then solve.
e.g. $\tfrac{x}{2} + \tfrac{x}{3} = 10$ , times 6: $3x + 2x = 60 \Rightarrow x = 12$
1.1.8
Solving Linear Equations Containing Binomials
Expand every product, including a negative sign in front of parentheses. If $x^2$ terms appear on both sides, they cancel and the equation becomes linear. Then collect $x$ terms and solve.
e.g. $(x+3)(x-2) = (x+1)(x+4)$ : $x^2 + x - 6 = x^2 + 5x + 4$ , so $-4x = 10$ and $x = -\tfrac52$
1.1.9
Solving Linear Equations Containing Binomials by Factoring
When the same binomial appears more than once, treat it as a single unit. Collect it on one side, solve for the binomial, then solve for $x$ .
e.g. $5(x-3) = 2(x-3) + 12 \Rightarrow 3(x-3) = 12 \Rightarrow x - 3 = 4 \Rightarrow x = 7$
$\displaystyle a(x+h) = b(x+h) + c$
$\displaystyle \Rightarrow\; (a-b)(x+h) = c$
1.1.10
Solving Linear Equations With Binomials on Both Sides
Distribute, combine like terms, then solve. If the variable cancels, a true statement means infinitely many solutions and a false one means no solution.
e.g. $3(x - 2) = 2(x + 4) \Rightarrow 3x - 6 = 2x + 8 \Rightarrow x = 14$
$\displaystyle a(x + b) = ax + ab$
1.1.11
Manipulating Linear Equations Containing Binomials
When asked for the value of an expression, solve for that expression directly instead of for $x$ . Then scale it to match what is asked.
e.g. $3(2x + 1) = 21 \Rightarrow 2x + 1 = 7$ , so $4x + 2 = 14$
$\displaystyle k(\text{expr}) = c \Rightarrow \text{expr} = \frac{c}{k}$

1.2 Word Problems and Modeling

1.2.1
Numerical Word Problems
Turn the words into one calculation. Keep track of units and check that the answer is reasonable.
e.g. A car goes 180 miles on 6 gallons. On 10 gallons it goes $\tfrac{180}{6} \times 10 = 300$ miles.
1.2.2
Multistep Numerical Word Problems
Break the problem into steps. Find each in-between quantity, label its units, and use it in the next step.
e.g. Adult tickets $8, kids $5. For 3 adults and 4 kids: $3(8) + 4(5) = 44$ , so change from $50 is $6.
1.2.3
Constructing Models With One-Variable Linear Equations
Let a variable stand for the unknown. The total is usually a starting amount plus a rate times the variable.
e.g. A gym charges a $25 fee plus $15 per month. A total of $100 gives $25 + 15m = 100$ .
$\displaystyle \text{total} = \text{start} + \text{rate} \cdot x$
1.2.4
Modeling and Solving Problems Using Linear Equations
Build the equation, then solve it with inverse operations. Answer the question that was asked, with units.
e.g. $25 + 15m = 100 \Rightarrow 15m = 75 \Rightarrow m = 5$ months
1.2.5
Linear Models Presented in Unsimplified Form
A model may come with parentheses or several separate terms. Distribute and combine like terms to reveal the rate and the starting value.
e.g. $C = 4(x + 3) + 2x = 6x + 12$ : $6 per item plus $12 fixed
1.2.6
Modeling Geometry Problems Using Linear Equations
Write each length or angle as an expression, then set the total equal to a known value such as a perimeter or an angle sum.
e.g. Length $w + 4$ , perimeter 28: $2(w + 4) + 2w = 28 \Rightarrow 4w + 8 = 28 \Rightarrow w = 5$
$\displaystyle P = 2l + 2w$
$\displaystyle \text{supplementary: } a + b = 180^\circ$
$\displaystyle \text{triangle: } a + b + c = 180^\circ$

1.3 Two-Variable Linear Equations

1.3.1
Solving Two-Variable Linear Models
Substitute the known value for one variable, then solve for the other.
e.g. $3x + 2y = 24$ with $x = 4$ : $12 + 2y = 24 \Rightarrow y = 6$
1.3.2
Constructing Models With Two-Variable Linear Equations
Give each kind of item its own variable. Multiply each variable by its price or rate, then add to get the total.
e.g. Pens cost $2 and notebooks $5. Spending $40 gives $2p + 5n = 40$ .
$\displaystyle ax + by = c$
1.3.3
Modeling Totals Using Two-Variable Linear Equations
Two kinds of item with a fixed total: price times count for each, added.
e.g. Adult tickets $12, child $8, $200 total: $12a + 8c = 200$
$\displaystyle ax + by = T$
1.3.4
Interpreting Two-Variable Linear Equations
Each term is a quantity in the context. A solution $(x, y)$ is a combination that makes the total come out exactly right.
e.g. In $2p + 5n = 40$ , the solution $(10, 4)$ means 10 pens and 4 notebooks cost exactly $40.
1.3.5
Interpreting Coefficients of Two-Variable Linear Equations
In $ax + by = c$ , $a$ is the value of one $x$ , $b$ is the value of one $y$ , and $c$ is the total. With the total fixed, raising $x$ by 1 changes $y$ by $-\tfrac{a}{b}$ .
e.g. In $2p + 5n = 40$ , each notebook costs $5. Buying 5 more pens means 2 fewer notebooks.
$\displaystyle \Delta y = -\frac{a}{b}\,\Delta x$

1.4 Linear Functions

1.4.1
Evaluating Linear Functions
Replace $x$ with the input everywhere, then simplify. Put negative inputs in parentheses.
e.g. $f(x) = 3x - 5$ : $f(-2) = 3(-2) - 5 = -11$
$\displaystyle f(x) = mx + b$
1.4.2
Finding a Function Input for a Given Output
Set the function equal to the given output and solve for $x$ .
e.g. $f(x) = 3x - 5 = 13 \Rightarrow 3x = 18 \Rightarrow x = 6$
$\displaystyle f(x) = k \;\Rightarrow\; \text{solve for } x$
1.4.3
Linear Functions Containing Unknown Parameters
Use a known input-output pair to find the unknown constant first. Then use the completed function.
e.g. $f(x) = kx + 4$ and $f(2) = 10$ : $2k + 4 = 10 \Rightarrow k = 3$ , so $f(5) = 19$
1.4.4
Working With Function Composition
$f(g(x))$ means apply $g$ first, then put that result into $f$ . Work from the inside out.
e.g. $f(x) = 2x + 1$ , $g(x) = x - 3$ : $f(g(5)) = f(2) = 5$
$\displaystyle (f \circ g)(x) = f(g(x))$

1.5 Functions in Tabular Form

1.5.1
Identifying Tables Corresponding to Linear Functions
A table is linear when the rate $\tfrac{\Delta y}{\Delta x}$ is the same for every pair of rows. With evenly spaced $x$ , the $y$ values change by the same amount each step.
e.g. $x$ : 0, 1, 2, 3 and $y$ : 4, 7, 10, 13 is linear, since $y$ rises by 3 each step.
$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$
1.5.2
Identifying Linear Functions Corresponding to Tables
Find the slope from two rows. Find $b$ from the row with $x = 0$ , or by substituting a point. On the SAT, plugging a row or two into each answer choice is often fastest.
e.g. Rows $(1, 5)$ and $(3, 11)$ : $m = 3$ , $5 = 3 + b \Rightarrow b = 2$ , so $y = 3x + 2$
$\displaystyle y = mx + b$
1.5.3
Linear Functions as Tables in Applications
In a context table, the slope is the rate per unit, and the value at $x = 0$ is the starting amount.
e.g. Hours 0, 2, 4 with cost $30, $70, $110: $20 per hour plus a $30 fee, so $C = 20h + 30$
1.5.4
Analyzing Functions in Tabular Form
To find $f(a)$ , find $a$ in the input row and read its output. To solve $f(x) = k$ , find $k$ in the output row and read its input.
e.g. If the table pairs $x = 4$ with $f(x) = 9$ , then $f(4) = 9$ , and $f(x) = 9$ when $x = 4$ .
1.5.5
Function Tables Containing Unknown Parameters
Substitute a complete row into the function to solve for the unknown constant. Then use the function to fill in missing entries.
e.g. $f(x) = ax - 3$ with row $(2, 7)$ : $2a - 3 = 7 \Rightarrow a = 5$ , so $f(4) = 17$

1.6 Graphs of Linear Equations

Pick the form by what you know: slope and intercept, slope and a point, or a quick way to both intercepts.

1.6.1
Interpreting Coefficients of Lines in Slope-Intercept Form
In $y = mx + b$ , $m$ is the slope, the change in $y$ for each 1-unit increase in $x$ . $b$ is the $y$ -intercept, the value of $y$ when $x = 0$ .
e.g. $y = -2x + 7$ : drops 2 for each unit right and crosses the $y$ -axis at $(0, 7)$
$\displaystyle y = mx + b$
1.6.2
Finding Equations of Lines Using Slopes and Points
Find the slope from two points if needed. Substitute one point into $y = mx + b$ to find $b$ , or use point-slope form.
e.g. Through $(2, 1)$ and $(4, 7)$ : $m = 3$ , $1 = 6 + b \Rightarrow b = -5$ , so $y = 3x - 5$
$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$
$\displaystyle y - y_1 = m(x - x_1)$
1.6.3
Graphs of Linear Equations in Slope-Intercept Form
Plot $(0, b)$ , then use the slope as rise over run to plot a second point. A positive slope rises left to right and a negative slope falls.
e.g. $y = \tfrac23 x - 1$ : start at $(0, -1)$ , go right 3 and up 2 to $(3, 1)$
1.6.4
Finding Intercepts of Lines in Slope-Intercept Form
The $y$ -intercept is $(0, b)$ . For the $x$ -intercept, set $y = 0$ and solve for $x$ .
e.g. $y = 2x - 8$ : $y$ -intercept $(0, -8)$ , $x$ -intercept $(4, 0)$
$\displaystyle 0 = mx + b$
$\displaystyle \Rightarrow\; x = -\frac{b}{m}$
1.6.5
Constructing and Analyzing Slope-Intercept Equations
Rewrite the equation as $y = mx + b$ by isolating $y$ . Then read the slope and $y$ -intercept directly.
e.g. $4x + 2y = 10 \Rightarrow 2y = -4x + 10 \Rightarrow y = -2x + 5$ : slope $-2$ , $y$ -intercept 5
1.6.6
Properties of Lines Given in Standard or General Form
For $Ax + By = C$ with $B \ne 0$ , the slope is $-\tfrac{A}{B}$ and the $y$ -intercept is $\tfrac{C}{B}$ . The $x$ -intercept is $\tfrac{C}{A}$ when $A \ne 0$ .
e.g. $3x + 4y = 12$ : slope $-\tfrac34$ , intercepts $(0, 3)$ and $(4, 0)$
$\displaystyle m = -\frac{A}{B}$
$\displaystyle y\text{-int} = \frac{C}{B},\quad x\text{-int} = \frac{C}{A}$
1.6.7
Analyzing Linear Equations in General Form
General form is $Ax + By + C = 0$ . Its slope is $-\tfrac{A}{B}$ and its $y$ -intercept is $-\tfrac{C}{B}$ . You can also just solve for $y$ .
e.g. $2x - y + 6 = 0 \Rightarrow y = 2x + 6$ : slope 2, $y$ -intercept 6
$\displaystyle m = -\frac{A}{B},\quad b = -\frac{C}{B}$
1.6.8
Analyzing Linear Equations of Shifted Lines
Shifting up $k$ adds $k$ to the equation and shifting down subtracts it. Shifting right $h$ replaces $x$ with $x - h$ , and shifting left replaces $x$ with $x + h$ . The slope never changes.
e.g. $y = 2x + 1$ shifted right 3: $y = 2(x - 3) + 1 = 2x - 5$
$\displaystyle \text{up } k\text{: } y = mx + b + k$
$\displaystyle \text{right } h\text{: } y = m(x - h) + b$
1.6.9
Constructing Linear Functions Using Slope-Intercept Form
Find the slope (the constant rate) and the $y$ -intercept (the starting value), then write $f(x) = mx + b$ .
e.g. Slope $-4$ through $(2, 1)$ : $1 = -8 + b \Rightarrow b = 9$ , so $f(x) = -4x + 9$
$\displaystyle f(x) = mx + b$

1.7 Parallel and Perpendicular Lines

The changes in x and y form a right triangle, so distance comes from the Pythagorean theorem. The midpoint averages the endpoints.

1.7.1
Parallel Lines
Parallel lines have equal slopes and different $y$ -intercepts. Equal slopes and equal intercepts give the same line.
e.g. Parallel to $y = 3x + 1$ through $(1, 5)$ : $5 = 3 + b$ , so $y = 3x + 2$
$\displaystyle m_1 = m_2$
1.7.2
Slopes of Perpendicular Lines
Perpendicular slopes are negative reciprocals: flip the fraction and change the sign. Their product is $-1$ . A horizontal line and a vertical line are also perpendicular.
e.g. Slope $\tfrac23$ has perpendicular slope $-\tfrac32$ .
$\displaystyle m_2 = -\frac{1}{m_1}$
$\displaystyle m_1 \cdot m_2 = -1$
1.7.3
Perpendicular Lines
Take the negative reciprocal of the given slope, then substitute the given point to find $b$ .
e.g. Perpendicular to $y = 2x - 3$ through $(4, 1)$ : $m = -\tfrac12$ , $1 = -2 + b$ , so $y = -\tfrac12 x + 3$

1.8 Modeling With Linear Functions

On a distance–time graph the slope is the speed. On a speed–time graph the slope is the acceleration, and the shaded area is the distance traveled.

1.8.1
Modeling With Linear Functions
A constant rate of change means a linear model, $f(x) = mx + b$ .
1.8.2
Constructing Models Using Linear Functions
A quantity that changes by a constant amount per unit is linear. The rate is $m$ and the initial value is $b$ .
e.g. A pool holds 200 gallons and fills at 30 gallons per minute: $V(t) = 30t + 200$
$\displaystyle f(x) = mx + b$
1.8.3
Further Constructing Models Using Linear Functions
When you get two data points instead of a rate, find the slope from the points, then solve for the intercept.
e.g. A plant is 5 in. at week 2 and 11 in. at week 5: $m = 2$ , so $h(t) = 2t + 1$
$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$
1.8.4
Interpreting Coefficients of Linear Functions in Context
The slope is the rate of change, in output units per input unit. The constant term is the output when the input is 0.
e.g. $C(m) = 0.25m + 40$ : $0.25 per mile, plus $40 before any miles are driven
1.8.5
Interpreting Graphs of Linear Functions
Read the slope as rise over run in context units. The $y$ -intercept is the starting value, and the $x$ -intercept is when the quantity reaches zero.
e.g. Gas in a tank, line from $(0, 12)$ to $(6, 0)$ : starts at 12 gallons, uses 2 gallons per hour, empty at 6 hours
1.8.6
Modeling Using Slope-Intercept Form
Write $y = mx + b$ with $m$ as the rate and $b$ as the starting amount. Then use it to predict values or solve for an input.
e.g. Savings of $50 plus $15 a week: $S = 15w + 50$ . For $200, $15w = 150$ , so $w = 10$ weeks.

1.9 Systems of Two-Variable Linear Equations

A system’s solution is where the lines meet. When elimination cancels both variables, the leftover statement tells you which case you have.

1.9.1
Solving Simple Systems of Linear Equations
A solution makes both equations true. Substitution: solve one equation for a variable and plug that expression into the other.
e.g. $y = 2x$ and $x + y = 12$ : $3x = 12$ , so $x = 4$ , $y = 8$
1.9.2
Identifying Structure in Systems of Linear Equations
Look for a shortcut before solving. Adding or subtracting the equations may give the requested expression directly.
e.g. $2x + y = 10$ and $x + 2y = 8$ : adding gives $3x + 3y = 18$ , so $x + y = 6$
1.9.3
Solving Systems of Equations Using Addition or Subtraction
When a variable has opposite coefficients, add the equations. When it has equal coefficients, subtract. Solve, then substitute back.
e.g. $x + y = 10$ and $x - y = 4$ : add to get $2x = 14$ , so $x = 7$ , $y = 3$
1.9.4
Solving Systems of Equations Using Elimination
Multiply one or both equations so a variable gets opposite coefficients, then add to eliminate it. Solve, then substitute back.
e.g. $2x + 3y = 12$ , $x - y = 1$ : times 3 gives $3x - 3y = 3$ . Add: $5x = 15$ , so $x = 3$ , $y = 2$
1.9.5
Modeling Word Problems With Systems of Linear Equations
Use one variable per unknown. Usually one equation counts the items and the other adds up their values.
e.g. 20 tickets, adults $10, kids $6, $164 total: $a + k = 20$ , $10a + 6k = 164 \Rightarrow a = 11$ , $k = 9$
$\displaystyle x + y = \text{count}$
$\displaystyle ax + by = \text{total value}$
1.9.6
Solving Problems With Systems of Linear Equations
Solve the system, then answer the actual question, which may ask for one variable, a sum, or a difference. On the SAT you can graph both lines in Desmos and read the intersection.
e.g. $y = x + 3$ , $y = -2x + 9$ : $3x = 6$ , so $x = 2$ , $y = 5$ . If asked for $x + y$ , answer 7.
1.9.7
Solving Problems With Systems of Linear Equations in Standard Form
With both equations in $Ax + By = C$ form, elimination is usually fastest. Line up like terms before adding or subtracting.
e.g. $3x + 2y = 16$ , $3x - y = 10$ : subtract to get $3y = 6$ , so $y = 2$ , $x = 4$
1.9.8
The Intersection of Two Lines
The intersection point is the solution of the system. Set the two $y$ -expressions equal to find $x$ , then find $y$ .
e.g. $y = x + 1$ and $y = -x + 5$ : $2x = 4$ , so they meet at $(2, 3)$
$\displaystyle m_1 x + b_1 = m_2 x + b_2$
1.9.9
The Relationship Between Systems of Linear Equations and Graphs
Different slopes: the lines cross once, one solution. Same slope, different intercepts: parallel, no solution. Same line: infinitely many solutions.
e.g. $2x + y = 3$ and $4x + 2y = 6$ are the same line, so there are infinitely many solutions.
$\displaystyle \tfrac{A_1}{A_2} = \tfrac{B_1}{B_2} \ne \tfrac{C_1}{C_2}\text{: none}$
$\displaystyle \tfrac{A_1}{A_2} = \tfrac{B_1}{B_2} = \tfrac{C_1}{C_2}\text{: infinite}$

1.10 Linear Equations With Infinitely Many or No Solutions

1.10.1
Determining the Number of Solutions to a Linear Equation
Simplify both sides to $ax + b = cx + d$ . Different $x$ -coefficients give exactly one solution. Same $x$ -coefficients give no solution if the constants differ and infinitely many if the constants match.
e.g. $3(x + 2) = 3x + 5 \Rightarrow 3x + 6 = 3x + 5$ : no solution
$\displaystyle ax + b = cx + d$
$\displaystyle a \ne c \Rightarrow \text{one solution}$
$\displaystyle a = c,\ b \ne d \Rightarrow \text{none}$
$\displaystyle a = c,\ b = d \Rightarrow \text{infinitely many}$
1.10.2
Linear Equations With Infinitely Many Solutions
The variables cancel and leave something always true. Every number works.
$\displaystyle 2(x + 1) = 2x + 2 \;\to\; 2 = 2$
1.10.3
Linear Equations With No Solutions
The variables cancel and leave something false. No number works.
$\displaystyle 3x + 1 = 3x + 5 \;\to\; 1 = 5$
1.10.4
Determining the Number of Solutions to a Linear System
Compare the two lines. Different slopes cross once: one solution. Same slope and different intercepts are parallel: no solution. The same line: infinitely many.
e.g. $y = 2x + 1$ and $y = 2x - 3$ : same slope, different intercepts, no solution
$\displaystyle \tfrac{a_1}{a_2} \ne \tfrac{b_1}{b_2} \Rightarrow \text{one}$
$\displaystyle \tfrac{a_1}{a_2} = \tfrac{b_1}{b_2} \ne \tfrac{c_1}{c_2} \Rightarrow \text{none}$
$\displaystyle \tfrac{a_1}{a_2} = \tfrac{b_1}{b_2} = \tfrac{c_1}{c_2} \Rightarrow \text{infinite}$
1.10.5
Linear Systems With Infinitely Many Solutions
Both equations describe the same line, so one equation is a constant multiple of the other. Every point on that line is a solution.
e.g. $2x + 3y = 6$ and $kx + 9y = 18$ : the second is 3 times the first, so $k = 6$
$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
1.10.6
Parametrizing Systems of Linear Equations
When a system has infinitely many solutions, describe all of them with a parameter. Set one variable equal to $t$ and solve for the other in terms of $t$ .
e.g. $x + 2y = 4$ : let $y = t$ , so the solutions are $(4 - 2t,\ t)$ . With $t = 1$ : $(2, 1)$
$\displaystyle ax + by = c$
$\displaystyle y = t \Rightarrow x = \frac{c - bt}{a}\;\; (a \ne 0)$
1.10.7
Linear Systems With No Solution
The lines are parallel: same slope, different $y$ -intercepts. In standard form the $x$ and $y$ coefficients are proportional but the constants are not.
e.g. $3x + ky = 5$ and $6x + 4y = 7$ : need $\tfrac36 = \tfrac{k}{4}$ , so $k = 2$ (and $\tfrac57 \ne \tfrac12$ )
$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$

1.11 Linear Inequalities One Variable

Closed dot and square bracket: the endpoint is included. Open dot and parenthesis: it is not. Infinity always takes a parenthesis.

1.11.1
Constructing Models With One-Variable Linear Inequalities
Turn the words into a sign. "At least" and "no less than" mean $\ge$ . "At most" and "no more than" mean $\le$ . "More than" means $>$ and "less than" means $<$ .
e.g. $20 plus $5 per GB, budget at most $60: $20 + 5g \le 60$
$\displaystyle \text{at least} \to \ge$
$\displaystyle \text{at most} \to \le$
1.11.2
Modeling and Solving With One-Variable Linear Inequalities
Write the inequality and solve it like an equation. Flip the sign when you multiply or divide by a negative. In context, round whole-number answers in the direction the inequality allows.
e.g. Tickets cost $12, need at least $100: $12n \ge 100 \Rightarrow n \ge 8.\overline{3}$ , so at least 9 tickets
$\displaystyle -2x < 6 \Rightarrow x > -3$
1.11.3
Modeling and Solving Percentage Problems With One-Variable Linear Inequalities
Write each percent as a decimal multiplier. A $p$ percent increase multiplies by $1 + r$ and a $p$ percent decrease multiplies by $1 - r$ , where $r = p/100$ .
e.g. After a 20 percent discount the price must be at most $48: $0.8p \le 48 \Rightarrow p \le 60$
$\displaystyle p\% \text{ of } x = \frac{p}{100}x$
$\displaystyle \text{increase: } x(1 + r)$
$\displaystyle \text{decrease: } x(1 - r)$
1.11.4
Modeling and Solving With One-Variable Linear Inequalities: Advanced Cases
For variables on both sides, such as comparing two plans, collect the variable terms on one side and the constants on the other. Flip the sign if you divide by a negative.
e.g. Plan A: $30 + $0.10/min; Plan B: $0.25/min. A is cheaper when $30 + 0.10m < 0.25m \Rightarrow m > 200$
$\displaystyle ax + b < cx + d$
$\displaystyle \Rightarrow\; (a - c)x < d - b$
1.11.5
Compound Inequalities
“And” means both are true, so $x$ is between. “Or” means at least one is true.
$\displaystyle -2 < x \le 5$
$\displaystyle x < -1 \;\text{or}\; x > 4$
1.11.6
Constructing Models With Compound Inequalities
A quantity between two bounds uses AND: $a \le x \le b$ . A quantity outside a range uses OR. "Within $d$ of $m$ " means $m - d \le x \le m + d$ .
e.g. Temperature within 3 degrees of 70: $67 \le T \le 73$
$\displaystyle a \le x \le b$
$\displaystyle \lvert x - m \rvert \le d$
$\displaystyle \Rightarrow\; m - d \le x \le m + d$

1.12 Linear Inequalities in Two Variables

Draw the boundary line, decide dashed or solid, then use a test point to pick the side to shade.

1.12.1
Solutions of Two-Variable Linear Inequalities
An ordered pair is a solution if substituting it makes the inequality true. All the solutions together form a half-plane.
e.g. Is $(2, 1)$ a solution of $3x - y > 4$ ? $6 - 1 = 5 > 4$ , yes
1.12.2
Modeling With Two-Variable Linear Inequalities
A budget or capacity limit becomes $ax + by \le C$ . Only sensible points count, often whole and nonnegative.
e.g. $60 for $5 books and $12 games: $5b + 12g \le 60$
1.12.3
Modeling and Solving Two-Variable Linear Inequalities
Write one inequality using both quantities, such as a total cost within a budget. Substitute a known value for one variable, then solve for the other.
e.g. Pens $2, notebooks $5, spend at most $30: $2p + 5n \le 30$ . With $n = 4$ : $2p \le 10$ , so $p \le 5$
$\displaystyle ax + by \le c$
1.12.4
Systems of Two-Variable Inequalities
The solution set is the overlap of the shaded half-planes. Use a solid boundary for $\le$ or $\ge$ and a dashed one for $<$ or $>$ . Shade above for $y >$ and below for $y <$ . A point must satisfy every inequality.
e.g. Is $(1, 2)$ in $y > x$ and $y \le 3$ ? $2 > 1$ and $2 \le 3$ , yes
1.12.5
Modeling With Systems of Inequalities
Write one inequality per constraint. Add $x \ge 0$ and $y \ge 0$ when the quantities cannot be negative.
e.g. Adult tickets $10, child $6, at least 50 tickets and at least $400: $a + c \ge 50$ , $10a + 6c \ge 400$
1.12.6
Modeling and Solving Systems of Linear Inequalities
Find the pairs that meet every constraint. To find the largest or smallest possible value, look where the boundary lines meet. On the SAT, testing the answer choices is often fastest.
e.g. $x + y \le 10$ and $y \ge 2x + 1$ : $2x + 1 \le 10 - x \Rightarrow x \le 3$ , so the largest $x$ is 3 (with $y = 7$ )

Unit 2 80 topics

Advanced Math

Polynomial Expressions, Rational Expressions, Exponential Expressions, and more.

2.13 Polynomial Expressions

A square with side a + b splits into four pieces. Its area is not a² + b²; the two rectangles make up the difference.

2.13.1
Adding Polynomials
Combine like terms: same variable raised to the same power. Add the coefficients and keep the exponent.
e.g. $(3x^2 + 2x - 5) + (x^2 - 4x + 1) = 4x^2 - 2x - 4$
$\displaystyle ax^n + bx^n = (a + b)x^n$
2.13.2
Adding Polynomials Using the Distributive Law
Distribute any coefficients first, then combine like terms. When subtracting a polynomial, distribute the negative to every term.
e.g. $2(x^2 + 3x) - (x^2 - x + 4) = x^2 + 7x - 4$
$\displaystyle a(b + c) = ab + ac$
$\displaystyle -(b - c) = -b + c$
2.13.3
Multiplying Binomials
Every term times every term: First, Outer, Inner, Last.
$\displaystyle (a + b)(c + d) = ac + ad + bc + bd$
$\displaystyle (x + 3)(x - 5) = x^2 - 2x - 15$
2.13.4
Factoring Polynomials Using the GCF
Factor out the greatest common factor: the largest number that divides every coefficient and the lowest power of each shared variable.
e.g. $6x^3 + 9x^2 = 3x^2(2x + 3)$
$\displaystyle ab + ac = a(b + c)$
2.13.5
Factoring Binomials Using Differences of Squares
A difference of two squares factors into a sum times a difference. A sum of squares $a^2 + b^2$ does not factor over the real numbers.
e.g. $4x^2 - 25 = (2x + 5)(2x - 5)$
$\displaystyle a^2 - b^2 = (a + b)(a - b)$
2.13.6
Factoring Trinomials
For $x^2 + bx + c$ , find two numbers that multiply to $c$ and add to $b$ .
e.g. $x^2 + 7x + 12 = (x + 3)(x + 4)$
$\displaystyle x^2 + bx + c = (x + p)(x + q)$
$\displaystyle pq = c,\quad p + q = b$
2.13.7
Factoring Higher-Order Polynomials
Take out the GCF first. Then look for a difference of squares, a quadratic pattern in $x^2$ , or grouping. Keep going until no factor can be factored further.
e.g. $x^3 + 3x^2 - 4x - 12 = x^2(x + 3) - 4(x + 3) = (x + 2)(x - 2)(x + 3)$
$\displaystyle x^4 - 16 = (x^2 + 4)(x^2 - 4)$
$\displaystyle = (x^2 + 4)(x + 2)(x - 2)$

2.14 Rational Expressions

2.14.1
Simplifying Rational Expressions by Factoring
Factor top and bottom completely, then cancel the shared factors.
$\displaystyle \frac{x^2 - 9}{x^2 + 3x} = \frac{(x - 3)(x + 3)}{x(x + 3)} = \frac{x - 3}{x}$
2.14.2
Simplifying Rational Expressions Using Polynomial Factorization
Factor the numerator and denominator fully, then cancel common factors, never single terms. Values that make the original denominator zero stay excluded.
e.g. $\tfrac{x^2 - 9}{x^2 + x - 6} = \tfrac{(x - 3)(x + 3)}{(x + 3)(x - 2)} = \tfrac{x - 3}{x - 2}$ , $x \ne -3, 2$
$\displaystyle \frac{ac}{bc} = \frac{a}{b} \quad (b, c \ne 0)$
2.14.3
Adding Rational Expressions
With the same denominator, add the numerators. With different denominators, rewrite both over a common denominator first.
e.g. $\tfrac{2}{x} + \tfrac{3}{x + 1} = \tfrac{2(x + 1) + 3x}{x(x + 1)} = \tfrac{5x + 2}{x(x + 1)}$
$\displaystyle \frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$
$\displaystyle \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$
2.14.4
Further Adding Rational Expressions
Factor each denominator, then use the least common denominator. When subtracting, distribute the negative across the whole numerator. Simplify, and remember values that make a denominator zero are excluded.
e.g. $\tfrac{1}{x - 2} + \tfrac{4}{x^2 - 4} = \tfrac{(x + 2) + 4}{(x - 2)(x + 2)} = \tfrac{x + 6}{x^2 - 4}$
$\displaystyle \frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}$
2.14.5
Fractions of Fractions
“Of” means multiply. Multiply across.
e.g. $\tfrac23$ of $\tfrac34$ $= \tfrac{6}{12} = \tfrac12$
$\displaystyle \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$

2.15 Exponential Expressions

2.15.1
Exponential Expressions
$a^n$ means $n$ factors of $a$ . Any nonzero base to the zero power is 1. A negative exponent means reciprocal. A fractional exponent means a root.
e.g. $8^{2/3} = (\sqrt[3]{8})^2 = 4$ and $2^{-3} = \tfrac18$
$\displaystyle a^0 = 1 \quad (a \ne 0)$
$\displaystyle a^{-n} = \frac{1}{a^n}$
$\displaystyle a^{m/n} = \left(\sqrt[n]{a}\right)^m \quad (a > 0)$
2.15.2
Simplifying Exponential Expressions
Same base: add exponents when multiplying and subtract when dividing. A power of a power multiplies exponents. A power of a product applies to each factor.
e.g. $\tfrac{(2x^3)^2 \cdot x}{4x^5} = \tfrac{4x^7}{4x^5} = x^2$
$\displaystyle a^m \cdot a^n = a^{m+n}$
$\displaystyle \frac{a^m}{a^n} = a^{m-n}$
$\displaystyle (a^m)^n = a^{mn}$
$\displaystyle (ab)^n = a^n b^n$
2.15.3
Simplifying Exponential Expressions Using the Distributive Law
Distribute a factor $a^m$ across a sum, adding exponents on matching bases. In reverse, factor out the smallest common power.
e.g. $2^{x+3} + 2^x = 2^x(2^3 + 1) = 9 \cdot 2^x$
$\displaystyle a^m(a^n + b) = a^{m+n} + a^m b$
$\displaystyle a^{n+k} + a^n = a^n(a^k + 1)$

2.16 Quadratic Equations

D = b² − 4ac. The solutions are where the parabola meets the x-axis, so D says whether it crosses twice, touches once, or misses.

2.16.1
Perfect Square Quadratic Equations
If one side is a perfect square, take the square root of both sides and include $\pm$ . If $k > 0$ there are two solutions, if $k = 0$ one, and if $k < 0$ no real solutions.
e.g. $(x - 3)^2 = 16 \Rightarrow x - 3 = \pm 4 \Rightarrow x = 7$ or $x = -1$
$\displaystyle a^2 + 2ab + b^2 = (a + b)^2$
$\displaystyle (x - h)^2 = k$
$\displaystyle \Rightarrow\; x = h \pm \sqrt{k}$
2.16.2
Solving Quadratic Equations Using the Zero-Product Rule
Move everything to one side so it equals 0, factor, and set each factor equal to 0. Never divide both sides by a variable, since that loses solutions.
e.g. $x^2 - 5x = 6 \Rightarrow (x - 6)(x + 1) = 0 \Rightarrow x = 6$ or $x = -1$
$\displaystyle ab = 0 \Rightarrow a = 0 \text{ or } b = 0$
2.16.3
Solving Quadratic Equations by Factoring
Get 0 on one side, factor, then set each factor to zero.
e.g. $x = 3$ or $-2$
$\displaystyle x^2 - x - 6 = 0$
$\displaystyle \Rightarrow\; (x - 3)(x + 2) = 0$
2.16.4
Solving Quadratic Equations With Leading Coefficients by Factoring
Factor $ax^2 + bx + c$ with the $ac$ method, then set each factor to zero.
e.g. $x = -\tfrac13$ or $2$
$\displaystyle 3x^2 - 5x - 2 = 0$
$\displaystyle \Rightarrow\; (3x + 1)(x - 2) = 0$
2.16.5
Solving Quadratic Equations Using the Quadratic Formula
Put the equation in the form $ax^2 + bx + c = 0$ first. The formula works for every quadratic. On the SAT, graphing in Desmos and reading the $x$ -intercepts is often faster.
e.g. $2x^2 + 3x - 2 = 0$ : $x = \tfrac{-3 \pm \sqrt{9 + 16}}{4} = \tfrac{-3 \pm 5}{4}$ , so $x = \tfrac12$ or $x = -2$
$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
2.16.6
Quadratic Equations With One Real Solution
Exactly one real solution when the discriminant is zero. The parabola touches the $x$ -axis at its vertex.
e.g. $x^2 + kx + 9 = 0$ has one solution when $k^2 - 36 = 0$ , so $k = \pm 6$
$\displaystyle b^2 - 4ac = 0$
$\displaystyle x = -\frac{b}{2a}$
2.16.7
Quadratic Equations With Two Real Solutions
Two different real solutions when the discriminant is positive. The parabola crosses the $x$ -axis twice.
e.g. $x^2 - 4x + c = 0$ : $16 - 4c > 0 \Rightarrow c < 4$
$\displaystyle b^2 - 4ac > 0$
2.16.8
Quadratic Equations With No Real Solutions
No real solutions when the discriminant is negative. The parabola never reaches the $x$ -axis.
e.g. $x^2 + 2x + 5 = 0$ : $4 - 20 = -16 < 0$ , no real solutions
$\displaystyle b^2 - 4ac < 0$
2.16.9
Counting Solutions to Quadratic Equations
The discriminant $D = b^2 - 4ac$ counts the real solutions. To count where a line meets a parabola, set them equal, rearrange to $ax^2 + bx + c = 0$ , and check $D$ .
e.g. $y = x^2$ and $y = 2x - 1$ : $x^2 - 2x + 1 = 0$ , $D = 4 - 4 = 0$ , one intersection point
$\displaystyle D = b^2 - 4ac$
$\displaystyle D > 0: 2 \quad D = 0: 1 \quad D < 0: 0$
2.16.10
Applying Vieta's Formulas
For $ax^2 + bx + c = 0$ with roots $r$ and $s$ , you can find the sum and product of the roots without solving.
e.g. $2x^2 - 6x + 3 = 0$ : sum of roots $3$ , product $\tfrac32$
$\displaystyle r + s = -\frac{b}{a}$
$\displaystyle rs = \frac{c}{a}$
2.16.11
Solving Quadratic Equations With Unknown Parameters
Treat the parameter as a number. Substitute a known root to find it, use the discriminant to control how many solutions there are, or use Vieta's formulas.
e.g. $x = 2$ solves $x^2 + kx - 10 = 0$ : $4 + 2k - 10 = 0 \Rightarrow k = 3$ . Product of roots is $-10$ , so the other root is $-5$

2.17 Quadratic Functions

The axis of symmetry runs through the vertex, halfway between the roots. Each form of the equation shows one of these features at a glance.

2.17.1
Quadratic Functions
A quadratic function has degree 2 and its graph is a parabola. It opens up if $a > 0$ and down if $a < 0$ . It can be written in standard, vertex, or factored form.
e.g. $x^2 - 2x - 3 = (x - 1)^2 - 4 = (x - 3)(x + 1)$
$\displaystyle f(x) = ax^2 + bx + c$
$\displaystyle f(x) = a(x - h)^2 + k$
$\displaystyle f(x) = a(x - r)(x - s)$
2.17.2
Graphs of Quadratic Functions
Key features: the vertex, the axis of symmetry through it, the $y$ -intercept $(0, c)$ , and the $x$ -intercepts at the real roots. A larger $\lvert a \rvert$ makes the parabola narrower.
e.g. $y = -x^2 + 4x$ : opens down, vertex $(2, 4)$ , $x$ -intercepts $0$ and $4$
$\displaystyle x = -\frac{b}{2a}$
$\displaystyle y\text{-intercept: } (0, c)$
2.17.3
Determining Inputs of Quadratic Functions for Given Outputs
Set $f(x)$ equal to the given output and solve the quadratic. There are usually two inputs, one if the output is the vertex value, and none if the output is beyond the vertex.
e.g. $f(x) = x^2 + 2x$ , $f(x) = 8$ : $(x + 4)(x - 2) = 0$ , so $x = -4$ or $x = 2$
$\displaystyle f(x) = k$
$\displaystyle \Rightarrow\; ax^2 + bx + (c - k) = 0$
2.17.4
Finding Extreme Values of Quadratic Functions in Factored Form
For $f(x) = a(x - r)(x - s)$ , the vertex is halfway between the zeros. Plug that $x$ in to get the extreme value: a minimum if $a > 0$ , a maximum if $a < 0$ .
e.g. $f(x) = (x - 1)(x - 5)$ : $x = 3$ , $f(3) = (2)(-2) = -4$ , minimum $-4$
$\displaystyle x = \frac{r + s}{2}$
2.17.5
Finding Extreme Values of Quadratic Functions in Vertex Form
In $f(x) = a(x - h)^2 + k$ the vertex is $(h, k)$ . The extreme value is $k$ , at $x = h$ . Watch the sign: $(x + 2)^2$ means $h = -2$ .
e.g. $f(x) = -2(x + 1)^2 + 7$ : maximum $7$ at $x = -1$
$\displaystyle f(x) = a(x - h)^2 + k$
$\displaystyle \text{vertex } (h, k)$
2.17.6
Finding Extreme Values of Quadratic Functions in Standard Form
Find the vertex $x$ -coordinate with $-\tfrac{b}{2a}$ , then substitute to get the extreme value. It is a minimum if $a > 0$ and a maximum if $a < 0$ .
e.g. $f(x) = 2x^2 - 8x + 3$ : $x = 2$ , $f(2) = 8 - 16 + 3 = -5$ , minimum $-5$
$\displaystyle x = -\frac{b}{2a}$
$\displaystyle \text{extreme} = f\!\left(-\frac{b}{2a}\right)$
2.17.7
Using Standard and Factored Forms to Reason About Quadratic Functions
Standard form shows the $y$ -intercept $c$ and the direction from $a$ . Factored form shows the $x$ -intercepts $r$ and $s$ . Expanding one form gives the other.
e.g. $(x - 2)(x + 6) = x^2 + 4x - 12$ : zeros $2$ and $-6$ , $y$ -intercept $-12$ , vertex at $x = -2$
$\displaystyle a(x - r)(x - s)$
$\displaystyle = ax^2 - a(r + s)x + ars$
2.17.8
Using Symmetry to Reason About Quadratic Functions
A parabola is symmetric about its axis. Two inputs with the same output are equally far from the axis, so the axis is at their midpoint.
e.g. $f(-1) = f(7)$ : axis $x = 3$ . If $f(0) = 9$ , then $f(6) = 9$
$\displaystyle f(x_1) = f(x_2)$
$\displaystyle \Rightarrow\; h = \frac{x_1 + x_2}{2}$
2.17.9
Identifying Quadratic Functions From Tabular Data
With equally spaced $x$ -values, constant first differences mean linear. Constant nonzero second differences mean quadratic. For an $x$ -step of 1, the second difference equals $2a$ .
e.g. $y$ : 1, 2, 5, 10 for $x$ = 0 to 3. First differences 1, 3, 5; second differences 2, 2, so quadratic with $a = 1$ : $y = x^2 + 1$
$\displaystyle \text{2nd difference} = 2a \quad (\Delta x = 1)$

2.18 Modeling With Quadratics

A thrown object’s height is a downward parabola in time. The vertex gives the peak; the positive root gives the landing.

2.18.1
Modeling Number Problems With Quadratic Equations
Turn the words into an equation, move everything to one side, and solve. Consecutive integers are $n$ and $n + 1$ ; consecutive even or odd integers are $n$ and $n + 2$ . Keep only answers that fit the conditions, such as "positive."
e.g. Two consecutive positive integers have product 56: $n(n + 1) = 56 \Rightarrow n^2 + n - 56 = 0 \Rightarrow (n + 8)(n - 7) = 0$ , so the integers are 7 and 8.
$\displaystyle \text{consecutive: } n,\; n + 1$
$\displaystyle \text{consecutive even/odd: } n,\; n + 2$
2.18.2
Modeling Geometry Problems With Quadratic Equations
Write each dimension in terms of one unknown, plug into the area or Pythagorean formula, and solve. Reject any answer that makes a length zero or negative.
e.g. Length is 3 more than width and area is 40: $w(w + 3) = 40 \Rightarrow w^2 + 3w - 40 = 0 \Rightarrow (w + 8)(w - 5) = 0$ , so $w = 5$ , $l = 8$ .
$\displaystyle A = lw$
$\displaystyle a^2 + b^2 = c^2$
2.18.3
Solving Border Problems With Quadratic Equations
A border of uniform width $x$ adds $2x$ to each dimension, since it sits on both sides. The border's own area is the total area minus the inner area.
e.g. A 6 by 4 photo plus frame has total area 48: $(6 + 2x)(4 + 2x) = 48 \Rightarrow 4x^2 + 20x - 24 = 0 \Rightarrow x^2 + 5x - 6 = 0$ , so $x = 1$ .
$\displaystyle (l + 2x)(w + 2x) = A_{\text{total}}$
$\displaystyle A_{\text{border}} = A_{\text{total}} - lw$
2.18.4
Interpreting Quadratic Functions in Modeling Contexts
Each form shows a different fact. Standard form: $c$ is the starting value at input 0. Vertex form: the max or min is $k$ at input $h$ . Factored form: the zeros are when the quantity equals 0.
e.g. $h(t) = -16(t - 2)^2 + 70$ : the ball reaches its maximum height of 70 ft at $t = 2$ s.
$\displaystyle f(x) = ax^2 + bx + c$
$\displaystyle f(x) = a(x - h)^2 + k$
$\displaystyle f(x) = a(x - r_1)(x - r_2)$
2.18.5
Interpreting Graphs of Quadratic Functions in Modeling Contexts
Read the vertex as the maximum or minimum and when it happens, the $y$ -intercept as the starting value, and the $x$ -intercepts as when the quantity is zero. Only the part of the graph that makes sense in context counts, such as $t \ge 0$ .
e.g. Height graph with $y$ -intercept $(0, 5)$ and vertex $(1.5, 41)$ : launched from 5 ft, highest point 41 ft at 1.5 s.
2.18.6
Modeling Planar Motion Using Quadratic Functions
Under gravity, height is a quadratic in time (in feet, $a = -16$ ; in meters, $a = -4.9$ ). The vertex gives the maximum height and when it occurs. The object lands when height is 0.
e.g. $h(t) = -16t^2 + 32t + 48$ : max at $t = 1$ , $h(1) = 64$ ft; $-16(t - 3)(t + 1) = 0$ , so it lands at $t = 3$ s.
$\displaystyle h(t) = -16t^2 + v_0 t + h_0$
$\displaystyle t_{\text{max}} = -\frac{b}{2a}$

2.19 Polynomials

2.19.1
Elementary Cubic Functions
The parent cubic $y = x^3$ passes through the origin, always increases, and is symmetric about the origin. In $y = a(x - h)^3 + k$ the center moves to $(h, k)$ , and $a < 0$ flips the graph.
e.g. $y = (x - 2)^3 + 1$ : center $(2, 1)$ ; it passes through $(3, 2)$ .
$\displaystyle y = a(x - h)^3 + k$
2.19.2
Higher-Degree Polynomials
The degree is the highest exponent and the leading coefficient multiplies that term. A polynomial of degree $n$ has at most $n$ real zeros and at most $n - 1$ turning points.
e.g. $p(x) = 2x^4 - 3x^2 + x - 5$ : degree 4, leading coefficient 2, constant term $-5$ .
$\displaystyle p(x) = a_n x^n + \dots + a_1 x + a_0$
2.19.3
Graphs of Higher-Degree Polynomials
Even degree: both ends point the same way (up if the leading coefficient is positive). Odd degree: the ends point opposite ways (rising on the right if positive). The graph crosses the $x$ -axis at a zero of odd multiplicity and touches and turns at a zero of even multiplicity.
e.g. $p(x) = -(x - 1)^2(x + 2)$ : up on the left, down on the right; touches at $x = 1$ , crosses at $x = -2$ ; $p(0) = -2$ .
$\displaystyle y\text{-intercept} = p(0)$
2.19.4
Polynomial Transformations
$f(x - h)$ shifts right $h$ and $f(x) + k$ shifts up $k$ . $-f(x)$ reflects over the $x$ -axis and $f(-x)$ reflects over the $y$ -axis. $a f(x)$ stretches vertically by $a$ .
e.g. Zeros of $f$ are $-1$ and $4$ , so zeros of $f(x - 3)$ are $2$ and $7$ .
$\displaystyle g(x) = a\,f(x - h) + k$
2.19.5
Further Polynomial Transformations
$f(bx)$ squeezes the graph horizontally by a factor of $b$ , so its zeros are the old zeros divided by $b$ . A vertical stretch $a f(x)$ leaves the zeros unchanged. Track a point through all the steps with the rule below.
e.g. Zeros of $f$ are $-2$ and $6$ , so zeros of $f(2x)$ are $-1$ and $3$ .
$\displaystyle g(x) = a\,f\big(b(x - h)\big) + k$
$\displaystyle (x, y) \to \left(\tfrac{x}{b} + h,\; ay + k\right)$
2.19.6
Using the Factor Theorem to Reason About Polynomials
$(x - c)$ is a factor of $p(x)$ exactly when $p(c) = 0$ . In general, $p(c)$ is the remainder when $p(x)$ is divided by $x - c$ .
e.g. $p(x) = x^3 - 2x^2 - 5x + 6$ : $p(1) = 1 - 2 - 5 + 6 = 0$ , so $x - 1$ is a factor.
$\displaystyle p(c) = 0 \iff (x - c)\text{ is a factor}$
$\displaystyle p(x) = (x - c)\,q(x) + p(c)$

2.20 Exponential Functions

Equal differences mean linear; equal ratios mean exponential. Any exponential growth eventually passes any linear growth.

2.20.1
Exponential Functions
Starting value $a$ , multiplied by base $b$ at each step. Growth if $b > 1$ , decay if $0 < b < 1$ .
$\displaystyle f(x) = a\,b^x \quad (a \ne 0,\ b > 0,\ b \ne 1)$
2.20.2
Maxima and Minima of Exponential Functions
An exponential function always increases or always decreases, so over all real numbers it has no maximum or minimum. If $a > 0$ its values stay above the asymptote $y = k$ without reaching it. On a closed interval the max and min are at the endpoints.
e.g. $f(x) = 3\cdot 2^x$ on $0 \le x \le 4$ : min $f(0) = 3$ , max $f(4) = 48$ .
$\displaystyle f(x) = a\cdot b^x + k$
$\displaystyle a > 0:\; f(x) > k$
2.20.3
Exponential Functions Containing Unknown Parameters
Use given points to find the unknowns. In $f(x) = a\cdot b^x$ , the value at $x = 0$ is $a$ ; then use a second point to solve for $b$ , keeping $b > 0$ .
e.g. $f(0) = 5$ and $f(2) = 45$ : $a = 5$ , $5b^2 = 45 \Rightarrow b = 3$ .
$\displaystyle f(x) = a\cdot b^x$
$\displaystyle f(0) = a$
2.20.4
Constructing Exponential Functions From Descriptions
Start with the initial value $a$ and multiply by the factor $b$ once per period. Growth of rate $r$ gives $b = 1 + r$ ; decay gives $b = 1 - r$ . If the change happens every $d$ units, use the exponent $t/d$ .
e.g. Starts at 200 and grows 5% per year: $f(t) = 200(1.05)^t$ .
$\displaystyle f(t) = a(1 \pm r)^t$
$\displaystyle f(t) = a\cdot 2^{t/d}\;\;(\text{doubles every } d)$
2.20.5
Exponential Functions Expressed in Equivalent Forms
Rewrite with exponent rules to show a different fact, such as the starting value or the rate per a different period.
e.g. $f(x) = 2^{x + 3} = 8\cdot 2^x$ , so $f(0) = 8$ . Also $100(1.1)^{2t} = 100(1.21)^t$ .
$\displaystyle b^{x + c} = b^c \cdot b^x$
$\displaystyle b^{kx} = \left(b^k\right)^x$
2.20.6
Transforming Exponential Functions
$a\cdot b^{x - h} + k$ shifts the graph right $h$ and up $k$ , moving the asymptote to $y = k$ . A negative $a$ flips the graph across the asymptote, so the values lie below $k$ .
e.g. $g(x) = 2^{x - 1} + 3$ : asymptote $y = 3$ , passes through $(1, 4)$ .
$\displaystyle g(x) = a\cdot b^{x - h} + k$
2.20.7
Identifying Exponential Functions From Tabular Data
With equally spaced $x$ -values, constant differences in $y$ mean linear and constant ratios mean exponential. The common ratio is $b$ and the value at $x = 0$ is $a$ .
e.g. $x = 0, 1, 2$ gives $y = 4, 12, 36$ : ratio 3, so $y = 4\cdot 3^x$ .
$\displaystyle \frac{f(x + 1)}{f(x)} = b$

2.21 Modeling With Exponential Functions

Adding 3 outside the exponential lifts the whole graph, including its asymptote. The y-intercept moves from (0, 1) to (0, 4).

2.21.1
Constructing Exponential Growth Models
Write the rate $r$ as a decimal and multiply by $1 + r$ each period. If the rate applies every $k$ units of time, the exponent is $t/k$ .
e.g. $1,000 at 4% per year: $A(t) = 1000(1.04)^t$ ; after 2 years, $1000(1.0816) = 1081.60$ , so $1,081.60.
$\displaystyle P(t) = P_0(1 + r)^t$
2.21.2
Constructing Exponential Decay Models
Losing $r$ per period means keeping $1 - r$ , so multiply by $1 - r$ each period. A half-life $h$ gives the factor $\tfrac12$ every $h$ units.
e.g. A $20,000 car loses 15% per year: $V(t) = 20000(0.85)^t$ ; after 2 years, $20000(0.7225) = 14450$ , so $14,450.
$\displaystyle P(t) = P_0(1 - r)^t$
$\displaystyle P(t) = P_0\left(\tfrac12\right)^{t/h}$
2.21.3
Determining Unknown Parameters in Exponential Models
Plug known data points into the model and solve. Dividing one equation by another cancels $a$ and leaves only $b$ .
e.g. $P(t) = ab^t$ with $P(2) = 50$ , $P(4) = 200$ : $b^2 = 4 \Rightarrow b = 2$ , then $4a = 50 \Rightarrow a = 12.5$ .
$\displaystyle \frac{a\,b^{t_2}}{a\,b^{t_1}} = b^{\,t_2 - t_1}$
2.21.4
Calculating Unknown Times in Exponential Models
Set the model equal to the target, isolate the power, and write both sides as powers of the same base. If the bases do not match, test the answer choices or find the intersection in Desmos.
e.g. $500\cdot 2^{t/3} = 4000 \Rightarrow 2^{t/3} = 8 = 2^3 \Rightarrow t = 9$ .
$\displaystyle a\cdot b^t = c$
$\displaystyle \Rightarrow\; b^t = \frac{c}{a}$
2.21.5
Converting Units of Time in Exponential Models
If the factor is per year and $t$ is in months, divide the exponent by 12. The equivalent factor per month is $b^{1/12}$ , and per decade it is $b^{10}$ .
e.g. Doubles every year, $t$ in months: $f(t) = a\cdot 2^{t/12}$ ; in 6 months the factor is $2^{1/2} \approx 1.41$ .
$\displaystyle f(t) = a\cdot b^{t/12}\;\;(t\text{ in months})$
$\displaystyle \text{monthly factor} = b^{1/12}$
2.21.6
Interpreting Exponential Models
In $f(t) = a\cdot b^{t/k}$ , $a$ is the starting value and $b$ is the factor applied every $k$ units. If $b > 1$ it grows by $b - 1$ per period; if $0 < b < 1$ it shrinks by $1 - b$ per period.
e.g. $f(t) = 300(0.92)^{t/2}$ : starts at 300 and decreases by 8% every 2 years.
$\displaystyle f(t) = a\cdot b^{t/k}$

2.22 Nonlinear Equations and Functions

2.22.1
Absolute Value Equations
Isolate the absolute value, then split into two equations. A negative right side has no solution.
e.g. $\lvert 2x - 1 \rvert = 7 \Rightarrow x = 4$ or $-3$
$\displaystyle \lvert A \rvert = k,\ k \ge 0$
$\displaystyle \Rightarrow\; A = k \text{ or } A = -k$
2.22.2
Rational Equations
Multiply both sides by the LCD to clear the fractions, then solve. Reject any answer that makes a denominator zero, since it is extraneous.
e.g. $\tfrac{3}{x} + \tfrac12 = 1$ : multiply by $2x$ : $6 + x = 2x \Rightarrow x = 6$ .
$\displaystyle \frac{a}{b} = \frac{c}{d} \;\Rightarrow\; ad = bc$
2.22.3
Radical Equations
Isolate the radical, then square both sides (cube for a cube root). Squaring can create extraneous solutions, so check every answer in the original equation.
e.g. $\sqrt{x + 3} = x - 3 \Rightarrow x^2 - 7x + 6 = 0$ , so $x = 1$ or $6$ . $x = 1$ gives $2 = -2$ (reject); $x = 6$ works.
$\displaystyle \sqrt{f(x)} = g(x)$
$\displaystyle \Rightarrow\; f(x) = g(x)^2,\; g(x) \ge 0$
2.22.4
Equations Containing Multiple Radicals
Isolate one radical and square. If a radical remains, isolate it and square again. Check every answer in the original equation.
e.g. $\sqrt{x + 5} = 1 + \sqrt{x} \Rightarrow x + 5 = 1 + 2\sqrt{x} + x \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4$ . Check: $3 = 1 + 2$ .
$\displaystyle \sqrt{a} = \sqrt{b} \;\Rightarrow\; a = b$
$\displaystyle (p + \sqrt{q})^2 = p^2 + 2p\sqrt{q} + q$
2.22.5
Absolute Value Functions
$f(x) = a\lvert x - h \rvert + k$ is a V with vertex $(h, k)$ and side slopes $\pm a$ . It opens up if $a > 0$ and down if $a < 0$ .
e.g. $f(x) = -2\lvert x - 1 \rvert + 4$ : vertex $(1, 4)$ , maximum 4, zeros at $x = -1$ and $x = 3$ .
$\displaystyle f(x) = a\lvert x - h \rvert + k$
$\displaystyle \lvert x \rvert = c \Rightarrow x = \pm c\;\;(c \ge 0)$
2.22.6
Rational Functions
A vertical asymptote sits at each zero of the denominator that does not cancel; a canceled factor leaves a hole. Horizontal asymptote: $y = 0$ if the bottom has higher degree, the ratio of leading coefficients if the degrees are equal, and none if the top has higher degree.
e.g. $f(x) = \frac{2x + 1}{x - 3}$ : vertical asymptote $x = 3$ , horizontal asymptote $y = 2$ .
$\displaystyle f(x) = \frac{p(x)}{q(x)}$

2.23 Multivariable Equations

2.23.1
Solving Multivariable Equations
To solve for one variable, treat every other letter as a constant and undo operations in reverse order.
e.g. Solve $P = 2l + 2w$ for $w$ : $P - 2l = 2w \Rightarrow w = \frac{P - 2l}{2}$ .
2.23.2
Further Multivariable Equations
If the target variable appears in more than one term, move those terms to one side, factor the variable out, and divide.
e.g. Solve $ax + b = cx$ for $x$ : $ax - cx = -b \Rightarrow x(a - c) = -b \Rightarrow x = \frac{b}{c - a}$ .
$\displaystyle ax - cx = x(a - c)$
2.23.3
Solving Multivariable Rational Equations
Multiply by the LCD to clear fractions, or combine fractions and flip both sides when each side is a single fraction. Then isolate the variable, factoring it out if it appears twice.
e.g. Solve $\tfrac1f = \tfrac1a + \tfrac1b$ for $f$ : $\tfrac1f = \tfrac{a + b}{ab} \Rightarrow f = \frac{ab}{a + b}$ .
$\displaystyle \frac{1}{x} = \frac{p}{q} \;\Rightarrow\; x = \frac{q}{p}$
2.23.4
Solving Multivariable Nonlinear Equations
Undo powers with roots and roots with powers. Taking an even root gives $\pm$ , unless the variable is known to be positive, such as a length.
e.g. Solve $T = 2\pi\sqrt{L/g}$ for $L$ : $\frac{T}{2\pi} = \sqrt{\tfrac{L}{g}} \Rightarrow \frac{T^2}{4\pi^2} = \frac{L}{g} \Rightarrow L = \frac{gT^2}{4\pi^2}$ .
$\displaystyle x^2 = c \;\Rightarrow\; x = \pm\sqrt{c}$
$\displaystyle \sqrt{x} = c \;\Rightarrow\; x = c^2\;\; (c \ge 0)$

2.24 Interpreting Graphs

2.24.1
Identifying Intercepts of Functions From Graphs
The $y$ -intercept is where the graph crosses the $y$ -axis, at $(0, f(0))$ ; a function has at most one. The $x$ -intercepts are where the graph meets the $x$ -axis, the zeros of $f$ .
e.g. Graph through $(-2, 0)$ , $(3, 0)$ , and $(0, -6)$ : zeros $-2$ and $3$ , and $f(0) = -6$ .
$\displaystyle y\text{-intercept: } (0,\, f(0))$
$\displaystyle x\text{-intercepts: } f(x) = 0$
2.24.2
Identifying Graph Properties in Modeling Contexts
Find the intercepts, the maximum and minimum points, and where the graph is increasing (rising left to right) or decreasing. Read each value with its units from the axis labels.
e.g. Temperature rises from 6 a.m. to 3 p.m., then falls: increasing from 6 a.m. to 3 p.m., maximum at 3 p.m.
2.24.3
Interpreting Graph Properties in Modeling Contexts
Say what each feature means in the story. The $y$ -intercept is the starting amount, an $x$ -intercept is when the quantity reaches zero, the slope is the rate of change, and a maximum is the greatest value and when it occurs.
e.g. A water tank graph has $y$ -intercept 500 and $x$ -intercept 25: the tank starts with 500 gallons and is empty after 25 minutes.

2.25 Systems of Nonlinear Equations

2.25.1
Solving Constant-Quadratic Systems of Equations
Set the quadratic equal to the constant and solve. A horizontal line meets a parabola 0, 1, or 2 times, depending on where it sits relative to the vertex.
e.g. $y = 5$ and $y = x^2 - 4$ : $x^2 = 9$ , so the solutions are $(3, 5)$ and $(-3, 5)$ .
$\displaystyle y = c,\;\; y = ax^2 + bx + d$
$\displaystyle \Rightarrow\; ax^2 + bx + d = c$
2.25.2
Solving Linear-Quadratic Systems of Equations
Substitute the linear expression for $y$ into the quadratic, set it to zero, and solve for $x$ . Plug each $x$ into the linear equation to get $y$ .
e.g. $y = x + 1$ and $y = x^2 - 1$ : $x^2 - x - 2 = 0 \Rightarrow x = 2$ or $-1$ ; points $(2, 3)$ and $(-1, 0)$ .
$\displaystyle y = mx + k,\;\; y = ax^2 + bx + c$
$\displaystyle \Rightarrow\; ax^2 + (b - m)x + (c - k) = 0$
2.25.3
Solving Challenging Linear-Quadratic Systems of Equations
If the linear equation is in standard form, first solve it for the easier variable. Substitute, simplify carefully, and if the question asks for only one value, find only that.
e.g. $2x + y = 7$ and $y = x^2 - 2x + 3$ : $x^2 - 2x + 3 = 7 - 2x \Rightarrow x^2 = 4$ ; points $(2, 3)$ and $(-2, 11)$ .
2.25.4
Exploiting Structure to Solve Nonlinear Systems of Equations
Look for a shortcut before substituting. Add or subtract the equations to cancel a term, substitute a whole expression at once, or use a factoring identity.
e.g. $x^2 + y^2 = 13$ and $x^2 - y^2 = 5$ : add to get $2x^2 = 18$ , so $x^2 = 9$ and $y^2 = 4$ .
$\displaystyle (x + y)^2 = x^2 + 2xy + y^2$
$\displaystyle x^2 - y^2 = (x + y)(x - y)$
2.25.5
Linear-Quadratic Systems With One Solution
Substitute to get one quadratic $ax^2 + bx + c = 0$ . The system has exactly one solution when the discriminant is 0, which means the line is tangent to the parabola.
e.g. $y = x^2$ and $y = 2x + k$ : $x^2 - 2x - k = 0$ ; one solution when $4 + 4k = 0$ , so $k = -1$ .
$\displaystyle b^2 - 4ac = 0$
2.25.6
Linear-Quadratic Systems With Zero or Two Solutions
Substitute to get one quadratic and check its discriminant. Positive means the line crosses the parabola twice; negative means they never meet.
e.g. $y = x^2 + 3$ and $y = 2x$ : $x^2 - 2x + 3 = 0$ , discriminant $4 - 12 = -8 < 0$ , so no solution.
$\displaystyle b^2 - 4ac > 0 \;\Rightarrow\; \text{two solutions}$
$\displaystyle b^2 - 4ac < 0 \;\Rightarrow\; \text{no solution}$
2.25.7
Other Systems of Nonlinear Equations
Use substitution or elimination, then check each answer in both original equations. The solutions are the intersection points, so graphing both in Desmos is often fastest on the SAT.
e.g. $y = \tfrac{6}{x}$ and $y = x + 1$ : $x(x + 1) = 6 \Rightarrow x^2 + x - 6 = 0$ , so $x = 2$ or $-3$ ; points $(2, 3)$ and $(-3, -2)$ .

Unit 3 60 topics

Geometry and Trigonometry

Lines, Angles, and Triangles, Area and Perimeter, Right Triangles, and more.

3.26 Lines, Angles, and Triangles

The angle numbers used in the proofs below. Highlighted: 4 and 5, a pair of alternate interior angles.

3.26.1
Corresponding Angles
Same position at each crossing of the transversal. Equal when the lines are parallel.
3.26.2
Consecutive Angles
Same side of the transversal, both inside the lines. Supplementary when the lines are parallel.
$\displaystyle \alpha + \beta = 180^\circ$
3.26.3
Vertical Angles
When two lines cross, the opposite angles are equal.
3.26.4
Problem-Solving With Corresponding Angles
When a transversal crosses two parallel lines, corresponding angles are equal. Alternate interior angles are also equal, and same-side interior angles add to $180^\circ$ . Set equal angles equal and solve.
e.g. Corresponding angles $(3x + 10)^\circ$ and $70^\circ$ : $3x + 10 = 70$ , so $x = 20$
$\displaystyle \text{corresponding: equal}$
$\displaystyle \text{alternate interior: equal}$
$\displaystyle \text{same-side interior: sum } 180^\circ$
3.26.5
Interior Angles of Triangles
The three inside angles always add to 180°.
$\displaystyle a + b + c = 180^\circ$
3.26.6
Interior Angles of Special Triangles
The angles of every triangle add to $180^\circ$ . An equilateral triangle has three $60^\circ$ angles. An isosceles triangle has equal base angles, and an isosceles right triangle has angles $45^\circ$ , $45^\circ$ , $90^\circ$ .
e.g. Isosceles with vertex angle $40^\circ$ : each base angle is $\tfrac{180 - 40}{2} = 70^\circ$
$\displaystyle A + B + C = 180^\circ$
3.26.7
Exterior Angles of Triangles
An outside angle equals the two inside angles not touching it.
$\displaystyle d = a + b$
3.26.8
Solving Problems With Multiple Triangles
Use the $180^\circ$ sum in each triangle, plus vertical angles (equal) and angles on a straight line (sum $180^\circ$ ) to pass information from one triangle to the next. Start with the triangle you know the most about.
e.g. Triangle 1 has $50^\circ$ and $60^\circ$ , so its third angle is $70^\circ$ . Its vertical angle in triangle 2 is $70^\circ$ . With an $80^\circ$ angle, the last angle is $30^\circ$ .
3.26.9
Interior and Exterior Angles in Connected Triangles
An exterior angle of a triangle equals the sum of the two remote (non-adjacent) interior angles. An exterior angle and its adjacent interior angle add to $180^\circ$ .
e.g. Remote interior angles $45^\circ$ and $65^\circ$ : exterior angle $110^\circ$ , adjacent interior angle $70^\circ$
$\displaystyle \text{exterior} = A + B$
$\displaystyle \text{exterior} + C = 180^\circ$

3.27 Area and Perimeter

Height is always measured straight up from the base, at a right angle. Never use the slanted side.

3.27.1
Areas of Rectangles and Squares
Length times width.
$\displaystyle A = lw$
$\displaystyle A = s^2$
3.27.2
Perimeters of Squares, Rectangles, and Triangles
Perimeter is the total length around a shape. Add all the side lengths.
e.g. Rectangle 7 by 3: $P = 2(7) + 2(3) = 20$
$\displaystyle P_{\text{square}} = 4s$
$\displaystyle P_{\text{rectangle}} = 2l + 2w$
$\displaystyle P_{\text{triangle}} = a + b + c$
3.27.3
Calculating Areas in the Coordinate Plane
Get lengths from coordinates. A horizontal segment's length is the difference of the $x$ -values, and a vertical segment's length is the difference of the $y$ -values. For a triangle, use a horizontal or vertical side as the base.
e.g. Triangle $(1, 1)$ , $(7, 1)$ , $(3, 5)$ : base $6$ , height $4$ , area $\tfrac12(6)(4) = 12$
$\displaystyle \text{horizontal} = \lvert x_2 - x_1 \rvert$
$\displaystyle \text{vertical} = \lvert y_2 - y_1 \rvert$
3.27.4
Constructing Functions Representing Area and Perimeter
Write every side in terms of one variable, then substitute into the area or perimeter formula.
e.g. Length is 3 more than width $w$ : $A(w) = w(w + 3)$ , $P(w) = 4w + 6$
$\displaystyle A = lw$
$\displaystyle P = 2l + 2w$

3.28 Right Triangles

Memorize these two shapes. Any 45°–45°–90° triangle has sides in the ratio 1 : 1 : √2; any 30°–60°–90° triangle, 1 : √3 : 2. The hypotenuse is always the longest side.

3.28.1
Areas of Right Triangles
The two legs are the base and height. If you know the hypotenuse and one leg, find the other leg with the Pythagorean theorem first.
e.g. Hypotenuse 13, leg 5: other leg $\sqrt{169 - 25} = 12$ , area $\tfrac12(5)(12) = 30$
$\displaystyle A = \tfrac12 ab$
3.28.2
The Pythagorean Theorem
In a right triangle, the square of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the legs. It works in reverse too: if the sides fit, the angle is right.
e.g. Legs 5 and 12 → $c = \sqrt{169} = 13$
$\displaystyle a^2 + b^2 = c^2$
3.28.3
Modeling Situations Using the Pythagorean Theorem
Look for a right angle: a ladder against a wall, walking north then east, the diagonal of a rectangle. The hypotenuse $c$ is the side opposite the right angle.
e.g. 10 ft ladder, foot 6 ft from the wall: top is $\sqrt{100 - 36} = 8$ ft up
$\displaystyle a^2 + b^2 = c^2$
3.28.4
Problem-Solving With Isosceles Right Triangles
A $45^\circ$ - $45^\circ$ - $90^\circ$ triangle has equal legs. The hypotenuse is a leg times $\sqrt2$ .
e.g. Hypotenuse 10: each leg is $\tfrac{10}{\sqrt2} = 5\sqrt2$ , area $\tfrac12(5\sqrt2)^2 = 25$
$\displaystyle x : x : x\sqrt2$
$\displaystyle \text{leg} = \frac{\text{hypotenuse}}{\sqrt2}$
3.28.5
Problem-Solving With Special Triangles
In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle the short leg is opposite $30^\circ$ , the hypotenuse is twice the short leg, and the long leg is the short leg times $\sqrt3$ . These ratios are on the SAT reference sheet.
e.g. Hypotenuse 12: short leg $6$ , long leg $6\sqrt3$
$\displaystyle 30^\circ\text{-}60^\circ\text{-}90^\circ:\; x : x\sqrt3 : 2x$
$\displaystyle 45^\circ\text{-}45^\circ\text{-}90^\circ:\; x : x : x\sqrt2$
3.28.6
Further Problem-Solving With Special Triangles
Split shapes into special triangles. The altitude of an equilateral triangle makes two $30^\circ$ - $60^\circ$ - $90^\circ$ triangles. The diagonal of a square makes two $45^\circ$ - $45^\circ$ - $90^\circ$ triangles.
e.g. Equilateral side 8: height $4\sqrt3$ , area $\tfrac12(8)(4\sqrt3) = 16\sqrt3$
$\displaystyle h = \frac{\sqrt3}{2}s \quad \text{(equilateral)}$
$\displaystyle d = s\sqrt2 \quad \text{(square)}$
3.28.7
Solving Right Triangle Problems by Manipulating Radicals
Simplify a radical by pulling out perfect-square factors. Rationalize a denominator by multiplying top and bottom by the radical.
e.g. Legs 4 and 8: $c = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt5$
$\displaystyle \sqrt{ab} = \sqrt a\,\sqrt b\;\; (a, b \ge 0)$
$\displaystyle \frac{a}{\sqrt b} = \frac{a\sqrt b}{b}$

3.29 Circles

An arc is part of the edge, a sector is a slice, and a segment is cut off by a chord. Arc length and sector area are the same fraction, θ/360°, of the whole.

3.29.1
Radians
A radian is the angle whose arc length equals the radius. A full circle is $2\pi$ radians, or $360^\circ$ .
e.g. $120^\circ = 120 \cdot \tfrac{\pi}{180} = \tfrac{2\pi}{3}$ radians
$\displaystyle \pi \text{ rad} = 180^\circ$
$\displaystyle \text{deg to rad: } \times \frac{\pi}{180}$
$\displaystyle \text{rad to deg: } \times \frac{180}{\pi}$
3.29.2
Arc Lengths of Sectors
An arc is the same fraction of the circumference as its central angle is of the full circle. With the angle in radians, arc length is radius times angle.
e.g. $r = 6$ , $\theta = 60^\circ$ : $s = \tfrac16 \cdot 12\pi = 2\pi$
$\displaystyle s = \frac{\theta}{360^\circ} \cdot 2\pi r$
$\displaystyle s = r\theta \quad (\theta \text{ in radians})$
$\displaystyle A_{\text{sector}} = \frac{\theta}{360^\circ} \cdot \pi r^2$
3.29.3
Areas and Circumferences of Circles
The radius is half the diameter. Check which one the problem gives before using a formula.
e.g. Diameter 10: $r = 5$ , $C = 10\pi$ , $A = 25\pi$
$\displaystyle A = \pi r^2$
$\displaystyle C = 2\pi r = \pi d$
3.29.4
Isosceles Triangles Formed Within Circles
Two radii to points on a circle form an isosceles triangle with the center. Its base angles are equal.
e.g. Central angle $80^\circ$ : each base angle is $50^\circ$
$\displaystyle \text{base angle} = \frac{180^\circ - \text{central}}{2}$
3.29.5
Tangent Lines to Circles
A tangent line touches the circle at exactly one point and is perpendicular to the radius at that point. This makes a right triangle with the center and any outside point on the tangent.
e.g. Radius 5, center to outside point 13: tangent length $= \sqrt{169 - 25} = 12$
$\displaystyle \overline{OP} \perp \text{tangent at } P$
$\displaystyle r^2 + t^2 = d^2$
3.29.6
Bisectors of Chords
A segment from the center perpendicular to a chord bisects the chord. The radius, half the chord, and the distance $d$ from the center form a right triangle.
e.g. Chord 16, radius 10: distance from center $\sqrt{100 - 64} = 6$
$\displaystyle r^2 = d^2 + \left(\tfrac{c}{2}\right)^2$

3.30 Circles in the Coordinate Plane

A circle is every point a fixed distance r from its center. The distance formula squared gives the equation, so the center is (2, 1) and the radius is 3.

3.30.1
The Canonical Equation of a Circle
A circle with center $(h, k)$ and radius $r$ . The signs inside the parentheses are opposite the center's coordinates.
e.g. $(x - 2)^2 + (y + 3)^2 = 25$ : center $(2, -3)$ , radius $5$
$\displaystyle (x - h)^2 + (y - k)^2 = r^2$
3.30.2
Canonical Equations of Transformed Circles
A translation moves the center and keeps the radius. Shifting right $a$ replaces $x$ with $x - a$ . Shifting up $b$ replaces $y$ with $y - b$ .
e.g. $x^2 + y^2 = 9$ shifted 4 left and 1 up: $(x + 4)^2 + (y - 1)^2 = 9$
$\displaystyle (h, k) \to (h + a,\, k + b)$
3.30.3
General Equations of Circles in the Coordinate Plane
Expanded form of a circle. Complete the square in $x$ and in $y$ to get the canonical form, adding the same amounts to both sides.
e.g. $x^2 + y^2 - 6x + 4y - 12 = 0 \Rightarrow (x - 3)^2 + (y + 2)^2 = 25$
$\displaystyle x^2 + y^2 + Dx + Ey + F = 0$
$\displaystyle \text{center } \left(-\tfrac{D}{2},\, -\tfrac{E}{2}\right)$
3.30.4
Solving Problems Involving the Canonical Equation of a Circle
A point is on the circle exactly when it satisfies the equation, so its distance from the center is $r$ . The center of a diameter is its midpoint.
e.g. Center $(1, 2)$ through $(4, 6)$ : $r^2 = 9 + 16 = 25$ , so $(x - 1)^2 + (y - 2)^2 = 25$
$\displaystyle r^2 = (x_1 - h)^2 + (y_1 - k)^2$

3.31 Inscribed Shapes

3.31.1
Solving Problems Involving Inscribed Squares
For a square inscribed in a circle, the square's diagonal is a diameter.
e.g. $r = 5$ : diagonal $10$ , side $5\sqrt2$ , area $50$
$\displaystyle s\sqrt2 = 2r$
$\displaystyle A_{\text{square}} = 2r^2$
3.31.2
Solving Problems Involving Inscribed Circles
For a circle inscribed in a square, the circle's diameter equals the square's side.
e.g. Square side 8: $r = 4$ , area outside the circle $64 - 16\pi$
$\displaystyle 2r = s$
3.31.3
Solving Problems Involving Circles Inscribed in Equilateral Triangles
The center of the circle is the triangle's center. The radius is one third of the triangle's height.
e.g. Side 6: height $3\sqrt3$ , radius $\sqrt3$
$\displaystyle r = \frac{h}{3} = \frac{s\sqrt3}{6}$
3.31.4
Solving Problems Involving Equilateral Triangles Inscribed in Circles
The circle's center is the triangle's center. The radius is two thirds of the triangle's height.
e.g. $R = 4$ : side $4\sqrt3$ , height $6$
$\displaystyle R = \frac{2h}{3} = \frac{s\sqrt3}{3}$
$\displaystyle s = R\sqrt3$

3.32 Surface Area and Volume

The face diagonal across the bottom and the height form a right triangle whose hypotenuse is the space diagonal. Using the Pythagorean theorem twice gives d = √(l² + w² + h²).

3.32.1
Volumes of Right Rectangular Prisms
Volume is length times width times height, in cubic units.
e.g. $2 \times 3 \times 5$ box: $V = 30$ cubic units
$\displaystyle V = lwh$
3.32.2
Surface Areas of Right Rectangular Prisms
Add the areas of all six faces. Opposite faces come in matching pairs.
e.g. $2 \times 3 \times 5$ box: $SA = 2(6 + 10 + 15) = 62$
$\displaystyle SA = 2(lw + lh + wh)$
3.32.3
Volumes and Surface Areas of Cubes
A cube has six identical square faces with side $s$ .
e.g. $s = 4$ : $V = 64$ , $SA = 96$
$\displaystyle V = s^3$
$\displaystyle SA = 6s^2$
3.32.4
Volumes of Cylinders
Base area times height, like a prism with a circular base.
e.g. $r = 3$ , $h = 5$ : $V = 45\pi \approx 141.4$
$\displaystyle V = \pi r^2 h$
3.32.5
Volumes of Cylinders: Word Problems
Volume is base area times height. Halve a given diameter to get $r$ , and keep units consistent.
e.g. Tank 6 ft across, 10 ft tall: $V = \pi(3)^2(10) = 90\pi \approx 283$ cubic feet
$\displaystyle V = \pi r^2 h$
3.32.6
Volumes of Spheres
A hemisphere has half the volume of the sphere.
e.g. $r = 3$ : $V = \tfrac43 \cdot 27\pi = 36\pi \approx 113.1$
$\displaystyle V = \frac{4}{3}\pi r^3$
3.32.7
Volumes of Cones
A cone holds one third of the cylinder with the same base and height. If you are given the slant height $\ell$ , find the height with $h = \sqrt{\ell^2 - r^2}$ .
e.g. $r = 3$ , $h = 4$ : $V = \tfrac13 \pi (9)(4) = 12\pi$
$\displaystyle V = \tfrac13 \pi r^2 h$
3.32.8
Volumes of Composite Solids
Break the solid into basic pieces. Add the volumes of the pieces, and subtract any part that is cut out.
e.g. Cylinder $r = 2$ , $h = 6$ with a cone $r = 2$ , $h = 3$ on top: $24\pi + 4\pi = 28\pi$
$\displaystyle V = V_1 + V_2 - V_{\text{removed}}$
3.32.9
Constructing Functions Representing Surface Area and Volume
Write every dimension in terms of one variable, then substitute into the volume or surface area formula.
e.g. Square base side $x$ , height $2x$ : $V(x) = 2x^3$ , $S(x) = 2x^2 + 8x^2 = 10x^2$
3.32.10
Surface Areas of Combined Rectangular Prisms
Add the surface areas of the separate prisms, then subtract twice the contact area, since those faces are hidden on both prisms.
e.g. Two unit cubes glued face to face: $6 + 6 - 2(1) = 10$
$\displaystyle S = S_1 + S_2 - 2A_{\text{contact}}$

3.33 Congruence and Similarity

Multiplying every side by k = 2 keeps the angles the same. Area grows by k², so the bigger triangle has 4 times the area.

3.33.1
Congruent Triangles
Congruent triangles have the same size and shape, so all corresponding sides and angles are equal. The letter order in $\triangle ABC \cong \triangle DEF$ shows which parts match. SSA and AAA do not prove congruence.
e.g. $\triangle ABC \cong \triangle DEF$ and $AB = 7$ : then $DE = 7$
$\displaystyle \text{SSS, SAS, ASA, AAS}$
$\displaystyle \text{HL (right triangles only)}$
3.33.2
Finding Corresponding Angles in Similar Shapes
Similar shapes have equal corresponding angles. Scaling changes lengths, never angles. Match vertices by their order in the similarity statement.
e.g. $\triangle ABC \sim \triangle XYZ$ and $\angle B = 70^\circ$ : then $\angle Y = 70^\circ$
3.33.3
Finding Corresponding Lengths in Similar Shapes
Corresponding lengths in similar shapes are proportional. Every pair has the same scale factor $k$ .
e.g. Sides 4 and 6; the side matching 4 is 10: $k = 2.5$ , so the other side is $15$
$\displaystyle \frac{a'}{a} = \frac{b'}{b} = k$
3.33.4
Solving Problems With Similar Triangles
Two equal angle pairs make triangles similar (AA). A line parallel to one side cuts off a smaller similar triangle. Set up a proportion with matching sides in the same positions.
e.g. A 6 ft person casts a 4 ft shadow and a tree casts a 20 ft shadow: tree $= 6 \cdot \tfrac{20}{4} = 30$ ft
$\displaystyle \frac{\text{small}_1}{\text{large}_1} = \frac{\text{small}_2}{\text{large}_2}$
3.33.5
Solving Problems With Embedded Right Triangles
A smaller right triangle inside a larger one that shares an acute angle is similar to it by AA. Match legs to legs and hypotenuse to hypotenuse, then use a proportion.
e.g. Legs 6 and 8. A segment parallel to the 6 leg, 4 units along the 8 leg from its acute end, has length $6 \cdot \tfrac48 = 3$
$\displaystyle \frac{\text{leg}_{\text{small}}}{\text{leg}_{\text{large}}}$
$\displaystyle = \frac{\text{hyp}_{\text{small}}}{\text{hyp}_{\text{large}}}$
3.33.6
Further Problem-Solving With Embedded Right Triangles
The altitude to the hypotenuse splits a right triangle into two smaller triangles, both similar to the original. If the altitude $h$ cuts hypotenuse $c$ into $p$ and $q$ , with leg $a$ next to $p$ and leg $b$ next to $q$ :
e.g. $p = 4$ , $q = 9$ : $h = 6$ , $c = 13$ , $a = \sqrt{52} = 2\sqrt{13}$
$\displaystyle h^2 = pq$
$\displaystyle a^2 = pc$
$\displaystyle b^2 = qc$
3.33.7
Identifying Information Needed to Determine Similarity
Triangles are similar by AA (two equal angle pairs), SAS (two sides in proportion and the included angles equal), or SSS (all three sides in proportion).
e.g. Triangles with sides 3, 4, 5 and 6, 8, 10 are similar by SSS with $k = 2$
$\displaystyle \text{AA}$
$\displaystyle \text{SAS: } \frac{a'}{a} = \frac{b'}{b},\; \angle C' = \angle C$
$\displaystyle \text{SSS: } \frac{a'}{a} = \frac{b'}{b} = \frac{c'}{c}$
3.33.8
Relating Length and Area in Similar Shapes
If lengths scale by $k$ , areas scale by $k^2$ . To go from an area ratio back to a length ratio, take the square root.
e.g. Length ratio $3 : 1$ and small area 5: large area $9 \cdot 5 = 45$
$\displaystyle \frac{A'}{A} = k^2$
3.33.9
Length, Surface Area, and Volume in Similar Solids
If lengths scale by $k$ , surface areas scale by $k^2$ and volumes by $k^3$ .
e.g. Volume ratio $27 : 64$ : length ratio $3 : 4$ , surface area ratio $9 : 16$
$\displaystyle \frac{S'}{S} = k^2$
$\displaystyle \frac{V'}{V} = k^3$

3.34 Trigonometry

Each trig ratio compares two sides of a right triangle, named from the angle θ. The hypotenuse is always the longest side, across from the right angle.

3.34.1
Trigonometric Ratios in Right Triangles
For an acute angle $\theta$ in a right triangle, each ratio compares two sides named from that angle (SOH CAH TOA). Only for right triangles.
e.g. Legs 3 and 4, hypotenuse 5, $\theta$ opposite 3: $\sin\theta = \tfrac35$ , $\cos\theta = \tfrac45$ , $\tan\theta = \tfrac34$
$\displaystyle \sin\theta = \frac{\text{opp}}{\text{hyp}}$
$\displaystyle \cos\theta = \frac{\text{adj}}{\text{hyp}}$
$\displaystyle \tan\theta = \frac{\text{opp}}{\text{adj}}$
3.34.2
Calculating Lengths and Areas Using Trigonometry
Pick the ratio that links the known side to the unknown side, then solve. For any triangle, the area is half the product of two sides times the sine of the angle between them.
e.g. Hypotenuse 10, angle $30^\circ$ : legs $10\sin 30^\circ = 5$ and $10\cos 30^\circ = 5\sqrt3$ , area $\tfrac12(5)(5\sqrt3) = \tfrac{25\sqrt3}{2}$
$\displaystyle \text{opp} = \text{hyp}\cdot\sin\theta$
$\displaystyle \text{adj} = \text{hyp}\cdot\cos\theta$
$\displaystyle A = \tfrac12 ab\sin C$
3.34.3
Calculating Trigonometric Ratios Using the Pythagorean Theorem
Given one ratio, label two sides of a right triangle, find the third side with the Pythagorean theorem, then read off any other ratio.
e.g. $\sin\theta = \tfrac{5}{13}$ : adj $= \sqrt{169 - 25} = 12$ , so $\cos\theta = \tfrac{12}{13}$ , $\tan\theta = \tfrac{5}{12}$
$\displaystyle \text{opp}^2 + \text{adj}^2 = \text{hyp}^2$
3.34.4
Cofunction Relationships and Identities
The two acute angles of a right triangle add to $90^\circ$ , so the sine of one equals the cosine of the other. If $\sin A = \cos B$ for acute angles, then $A + B = 90^\circ$ .
e.g. $\sin(x + 10)^\circ = \cos(2x + 20)^\circ \Rightarrow 3x + 30 = 90 \Rightarrow x = 20$
$\displaystyle \sin\theta = \cos(90^\circ - \theta)$
$\displaystyle \cos\theta = \sin(90^\circ - \theta)$
$\displaystyle \tan\theta = \frac{1}{\tan(90^\circ - \theta)}$
3.34.5
Relations Between Trigonometric Ratios in Similar Right Triangles
Similar right triangles have equal corresponding angles, so their trig ratios are equal. A trig ratio depends only on the angle, not on the size of the triangle.
e.g. Triangles 3-4-5 and 6-8-10: $\sin A = \tfrac35 = \tfrac{6}{10}$
$\displaystyle \sin A = \frac{a}{c} = \frac{a'}{c'}$
3.34.6
Problem-Solving Using Trigonometric Ratios in Similar Right Triangles
Find the ratio in one triangle, then use it with a known side of the similar triangle to find the missing side. Match corresponding angles first.
e.g. $\tan A = \tfrac{5}{12}$ , $D$ corresponds to $A$ , side opposite $D$ is 15: adjacent $= 36$ , hypotenuse $= 39$
$\displaystyle \tan D = \tan A$
$\displaystyle \Rightarrow\; \frac{\text{opp}}{\text{adj}} \text{ is the same}$
3.34.7
Trigonometric Ratios of Large Special Angles
For angles past $90^\circ$ , use the reference angle (the acute angle to the $x$ -axis), then set the sign by quadrant. Positive ratios: all in QI, sin in QII, tan in QIII, cos in QIV.
e.g. $\cos 120^\circ = -\cos 60^\circ = -\tfrac12$ , $\sin 135^\circ = \tfrac{\sqrt2}{2}$
$\displaystyle \sin(180^\circ - \theta) = \sin\theta$
$\displaystyle \cos(180^\circ - \theta) = -\cos\theta$
$\displaystyle \tan(180^\circ + \theta) = \tan\theta$

Unit 4 67 topics

Problem-Solving and Data Analysis

Sampling and Surveys, Measures of Centrality, Representing Data, and more.

4.35 Sampling and Surveys

4.35.1
Sampling Bias and Validity
Results generalize only to the population the sample was randomly selected from. Convenience or voluntary-response samples are biased. A cause-and-effect conclusion needs random assignment to treatments.
e.g. Surveying gym members about exercise says nothing reliable about all town residents.
4.35.2
Estimation From Proportions and Samples
If the sample is random, apply the sample proportion to the whole population.
e.g. 18 of 60 sampled students bike; school of 1,200: $0.3 \times 1200 = 360$
$\displaystyle \text{estimate} = \frac{\text{count}}{\text{sample size}} \times N$
4.35.3
Interpreting Survey Data
A random sample's percentage estimates the percentage for the population it was drawn from, and no wider group. The estimate is approximate, not exact.
e.g. 120 of 400 randomly chosen voters support a measure: about $30\%$ of all voters do.
4.35.4
Margin of Error
The estimate plus or minus the margin of error gives a range of plausible values for the population value. A larger sample gives a smaller margin of error.
e.g. $52\% \pm 3\%$ : plausible range $49\%$ to $55\%$
$\displaystyle \text{estimate} - \text{MOE} \text{ to } \text{estimate} + \text{MOE}$
4.35.5
Interpreting Margin of Error
The interval describes the population mean or proportion, not individual values. It is plausible, not guaranteed. A larger sample shrinks it, but it does not correct bias.
e.g. Mean sleep 6.2 h with margin 0.4 h: the population mean is plausibly 5.8 to 6.6 h. It does not say every student sleeps that long.

4.36 Measures of Centrality

In a right-skewed histogram the few large values in the tail drag the mean to the right of the median. Here the mean is estimated from bin midpoints; the median is less affected by the tail.

4.36.1
The Mean of a Dataset
Add all values and divide by how many there are. Outliers pull the mean toward them.
e.g. $4, 7, 9, 12$ : $\tfrac{32}{4} = 8$
$\displaystyle \bar{x} = \frac{\text{sum of values}}{n}$
4.36.2
The Median of a Dataset
Sort the values. The median is the middle one, or the average of the two middle values when $n$ is even. Outliers barely affect it.
e.g. $3, 8, 5, 10 \to 3, 5, 8, 10$ : median $\tfrac{5 + 8}{2} = 6.5$
$\displaystyle \text{position} = \frac{n + 1}{2}$
4.36.3
Comparing Measures of Center
Symmetric with no outliers: use the mean. Skewed or with outliers: use the median.
4.36.4
The Relationship Between Means and Sums
The sum equals the mean times the number of values. Work with sums when values are added or removed, then divide again.
e.g. 5 tests average 82, so sum 410. A 94 makes $\tfrac{504}{6} = 84$ .
$\displaystyle \text{sum} = \bar{x} \cdot n$
4.36.5
Finding Unknown Observations From Centrality Measures
Turn the mean into a total, then subtract the known values. For a median, place the unknown so it lands in the right sorted position.
e.g. $70, 80, 85, x$ with mean 80: $x = 320 - 235 = 85$
$\displaystyle x = n\bar{x} - (\text{sum of others})$

4.37 Representing Data

Box plot. Each of the four pieces (whisker, half-box, half-box, whisker) holds about a quarter of the data.

4.37.1
Tabular Datasets
Find the row and column that match the question, and check whether it asks for a cell, a row or column total, or the grand total. Use the right total as the denominator for any fraction or percent.
e.g. 45 of 150 students chose soccer: $\tfrac{45}{150} = 0.3$ , or 30 percent
4.37.2
Frequency Tables
Each value (or interval) with a count of how often it occurs. The counts add up to $n$ .
4.37.3
Bar Graphs
Each bar's height is the value or frequency for its category. Read the scale carefully and check whether the axis starts at 0.
e.g. Bars of 12, 18, and 30: total 60, so the third category is $\tfrac{30}{60} = \tfrac12$ of the data
4.37.4
Box Plots
Draw the five-number summary. The box spans Q1 to Q3 with a line at the median; whiskers reach min and max.
$\displaystyle \min,\ Q_1,\ \text{median},\ Q_3,\ \max$
4.37.5
Measuring Centrality From Frequency Tables
Mean: multiply each value by its frequency, add, and divide by the total frequency. Median: count to position $\tfrac{n+1}{2}$ using running totals.
e.g. Values 1, 2, 3 with frequencies 2, 5, 3: mean $\tfrac{2 + 10 + 9}{10} = 2.1$ ; 5th and 6th values are 2, so median 2
$\displaystyle \bar{x} = \frac{\sum f x}{\sum f}$
4.37.6
Bounding Statistical Metrics
When data is grouped into intervals, the exact mean or range is unknown. Put every value at the low end of its interval for the minimum, and at the high end for the maximum.
e.g. 3 whole numbers in 10 to 19, 2 in 20 to 29: mean from $\tfrac{30 + 40}{5} = 14$ to $\tfrac{57 + 58}{5} = 23$
$\displaystyle \bar{x}_{\min} = \frac{\sum f\cdot\text{low}}{n}$
$\displaystyle \bar{x}_{\max} = \frac{\sum f\cdot\text{high}}{n}$

4.38 Comparing Data

Two box plots on one scale make comparison quick: the median line shows the center, and the width of the box (the IQR) shows how spread out the middle half is.

4.38.1
Comparing Data Sets
Compare center (mean or median) and spread (range or standard deviation). A larger standard deviation means values sit farther from the mean on average.
e.g. $\{5, 5, 5\}$ and $\{1, 5, 9\}$ : same mean 5, but the second has range 8 and a larger standard deviation
4.38.2
Comparing Data Sets Presented Visually
In dot plots and histograms, the center is where the data balances and the spread is how wide it lies. Data bunched near the center has a smaller standard deviation. In a box plot, the line is the median and the box width is the IQR.
e.g. Two dot plots centered at 10: the one with points packed tighter has the smaller standard deviation.
4.38.3
Comparing Centrality Measures From Frequency Tables
Find the mean or median of each table separately, then compare. Look at where the large frequencies sit, since they pull the center toward them.
e.g. Values 1, 2, 3. Table A, frequencies 3, 2, 1: median 1.5. Table B, frequencies 1, 2, 3: median 2.5.
4.38.4
Comparing Centrality Measures With Symmetric Datasets
In a symmetric dataset, mean equals median. A long right tail (skewed right) pulls the mean above the median. A long left tail pulls it below.
e.g. $1, 3, 5$ : mean = median = 3. $1, 3, 20$ : median 3, mean 8.
$\displaystyle \text{symmetric: } \bar{x} = \text{median}$
$\displaystyle \text{skewed right: } \bar{x} > \text{median}$
4.38.5
Comparing Statistical Effects of Data Change
Adding $c$ to every value shifts the mean and median by $c$ but leaves range and standard deviation unchanged. Multiplying every value by $k > 0$ multiplies all four by $k$ . Changing an extreme value moves the mean, range, and SD, but usually not the median.
e.g. $70, 80, 90$ , each $+5$ : mean $80 \to 85$ , range stays 20
$\displaystyle x + c:\ \text{center} + c,\ \text{spread same}$
$\displaystyle kx:\ \text{center and spread} \times k$
4.38.6
Effects of Additional Data on Centrality Measures
A new value above the mean raises the mean, below lowers it, and equal leaves it unchanged. The median moves at most to a neighboring value. A value far from the center increases the standard deviation.
e.g. $2, 4, 6, 8$ (mean 5, median 5), add 20: mean $\tfrac{40}{5} = 8$ , median 6

4.39 Probability

A ∪ B is everything in either circle, and A ∩ B is the shaded overlap. Adding P(A) and P(B) counts the overlap twice, so subtract it once.

4.39.1
Simple Probabilities in Real-World Contexts
With equally likely outcomes, probability is favorable outcomes over total outcomes. It is always from 0 to 1.
e.g. 3 red, 5 blue, 2 green: $P(\text{blue}) = \tfrac{5}{10} = \tfrac12$
$\displaystyle P(A) = \frac{\text{favorable}}{\text{total}}$
$\displaystyle P(\text{not } A) = 1 - P(A)$
4.39.2
Calculating Probabilities From Datasets
Count the data values that meet the condition and divide by the total number of values.
e.g. Rain on 10 of 25 recorded days: $P(\text{rain}) = \tfrac{10}{25} = 0.4$
$\displaystyle P = \frac{\text{matching values}}{\text{total values}}$
4.39.3
Calculating Probabilities From Frequency Tables
Add the frequencies of the categories that meet the condition and divide by the total frequency.
e.g. Shoe sizes 7, 8, 9 with frequencies 4, 10, 6: $P(\text{size} \ge 8) = \tfrac{16}{20} = 0.8$
$\displaystyle P = \frac{\text{matching frequency}}{\text{total frequency}}$
4.39.4
Simple Probabilities From Two-Way Frequency Tables
With no condition, the denominator is the grand total. Use a cell for "A and B" and a row or column total for just one category.
e.g. 100 students, 65 say yes, 30 are 9th graders who say yes: $P(\text{yes}) = 0.65$ , $P(\text{9th and yes}) = 0.3$
$\displaystyle P(A) = \frac{\text{total for } A}{\text{grand total}}$
4.39.5
Conditional Probabilities From Two-Way Frequency Tables
"Given $B$ " restricts you to the row or column for $B$ . That total becomes the denominator.
e.g. 50 9th graders, 30 say yes, 65 yes overall: $P(\text{yes} \mid \text{9th}) = \tfrac{30}{50} = 0.6$ , $P(\text{9th} \mid \text{yes}) = \tfrac{30}{65} = \tfrac{6}{13}$
$\displaystyle P(A \mid B) = \frac{\text{count of } A \text{ and } B}{\text{count of } B}$
4.39.6
Calculating Conditional Probabilities
Divide the probability of both events by the probability of the given event. Order matters: $P(A \mid B)$ usually differs from $P(B \mid A)$ .
e.g. $P(A \cap B) = 0.12$ , $P(B) = 0.4$ : $P(A \mid B) = 0.3$
$\displaystyle P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

4.40 Linear 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.

4.40.1
Making Predictions Using Best-Fit Lines
Substitute the $x$ -value into the line's equation to get the predicted $y$ . Predictions far outside the range of the data are unreliable.
e.g. $\hat{y} = 2.5x + 10$ at $x = 8$ : $\hat{y} = 30$
$\displaystyle \hat{y} = mx + b$
4.40.2
The Slope of a Best-Fit Line
The slope is the predicted change in $y$ for each 1-unit increase in $x$ . Use two points on the line, not data points, to compute it.
e.g. $\text{price} = 0.75w + 2$ : each extra pound adds a predicted $0.75
$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$
4.40.3
Identifying Best-Fit Lines
A best-fit line follows the trend, with points scattered about evenly above and below it. Check that its slope sign matches the direction and its intercept and one point match the graph.
e.g. Points rising from about $(0, 2)$ to $(10, 22)$ : best fit about $y = 2x + 2$
4.40.4
Best-Fit Lines With Compressed Axes
Axes may not start at 0 or may skip a section. Read the tick values when computing slope, and do not take the left edge of the graph as the $y$ -intercept.
e.g. Line through $(2000, 50)$ and $(2010, 80)$ : slope 3 per year; with $x$ = years since 2000, $y = 3x + 50$
4.40.5
Analyzing Residuals of Best-Fit Lines
A residual is actual minus predicted. Positive means the point is above the line, negative means below.
e.g. $\hat{y} = 3x + 1$ , point $(4, 15)$ : predicted 13, residual 2
$\displaystyle \text{residual} = y - \hat{y}$
4.40.6
Transformations of Best-Fit Lines
If every $y$ -value shifts by $c$ , the line shifts by $c$ and the slope stays the same. If every $y$ -value is multiplied by $k$ , both slope and intercept are multiplied by $k$ .
e.g. $\hat{y} = 2x + 5$ in dollars becomes $\hat{y} = 200x + 500$ in cents
$\displaystyle y \to y + c:\ \hat{y} = mx + (b + c)$
$\displaystyle y \to ky:\ \hat{y} = kmx + kb$

4.41 Nonlinear Data Models

4.41.1
Interpreting Data Plots
Describe a scatterplot by direction (positive or negative), form (linear or curved), strength, and outliers. An association does not prove cause and effect.
e.g. Points rise steeply and then level off: a positive, nonlinear association.
4.41.2
Interpreting Time-Series Graphs
Time runs along the $x$ -axis. Read the highs, lows, and where the graph rises or falls. The steepest segment shows the fastest change.
e.g. $60^\circ\text{F}$ at 8 AM to $75^\circ\text{F}$ at 1 PM: up $15^\circ$ in 5 h, or $3^\circ$ per hour
4.41.3
Average Rate of Change From Data Plots
The average rate of change between two points is the slope of the line through them, in $y$ -units per $x$ -unit.
e.g. Population 400 in 2010 and 700 in 2020: $\tfrac{300}{10} = 30$ per year
$\displaystyle \frac{f(b) - f(a)}{b - a}$
4.41.4
Linear vs. Exponential Models
A linear model adds the same amount for each equal step in $x$ . An exponential model multiplies by the same factor. Check for constant differences or constant ratios.
e.g. $3, 6, 9, 12$ : difference 3, linear. $3, 6, 12, 24$ : ratio 2, exponential.
$\displaystyle y = mx + b$
$\displaystyle y = a \cdot b^x$
4.41.5
Quadratic vs. Exponential Models
A quadratic has constant second differences and a turning point. An exponential has constant ratios, always rises or always falls, and levels off toward a horizontal asymptote on one side.
e.g. $1, 4, 9, 16$ : second differences 2, quadratic. $2, 6, 18, 54$ : ratio 3, exponential.
$\displaystyle y = ax^2 + bx + c$
$\displaystyle y = a \cdot b^x$

4.42 Percentages

Percent bar. The whole is always 100%. Line the percents up with the amounts: 25% of 80 is 20.

4.42.1
Finding Parts of Numbers From Percentages
To find a percent of a number, change the percent to a decimal and multiply by the whole. "Of" means multiply.
e.g. $30\%$ of $80 = 0.3 \times 80 = 24$
$\displaystyle \text{part} = \frac{p}{100} \times \text{whole}$
4.42.2
Expressing Ratios of Numbers as Percentages
Divide the part by the whole, then multiply by 100 to write it as a percent.
e.g. 18 out of 24: $\tfrac{18}{24} = 0.75 = 75\%$
$\displaystyle \text{percent} = \frac{\text{part}}{\text{whole}} \times 100\%$
4.42.3
Finding Percentages Represented by Variables
Treat variables like numbers: divide part by whole and multiply by 100. To turn " $p$ percent" into math, write $\frac{p}{100}$ .
e.g. If $20\%$ of $x$ is $y$ , then $0.2x = y$ , so $x = 5y$ : $x$ is $500\%$ of $y$
$\displaystyle x \text{ is } \frac{100x}{y}\% \text{ of } y$
$\displaystyle p\% \text{ of } q = \frac{pq}{100}$
4.42.4
Finding Totals From Percentages
If you know a part and the percent it represents, divide the part by the percent written as a decimal.
e.g. 15 is $25\%$ of what? $15 \div 0.25 = 60$
$\displaystyle \text{whole} = \frac{\text{part}}{p/100}$
4.42.5
Applying Percentages in Succession
Multiply the multipliers one after another. Never just add the percents.
e.g. +10% then −10% → $1.1 \times 0.9 = 0.99$ → 1% less than you started
$\displaystyle \text{new} = \text{original} \times m_1 \times m_2$

4.43 Percentage Change

Percent change is a multiplier. Up 25% means × 1.25; down 25% means × 0.75. To find the original, divide.

4.43.1
Expressing Percentage Change Using Decimals
A percent increase or decrease is one multiplication. Add the rate to 1 for an increase. Subtract it from 1 for a decrease.
e.g. Up $15\%$ : $\times 1.15$ . Down $8\%$ : $\times 0.92$
$\displaystyle \text{up } p\%: \;\times\left(1 + \tfrac{p}{100}\right)$
$\displaystyle \text{down } p\%: \;\times\left(1 - \tfrac{p}{100}\right)$
4.43.2
Calculating Percentage Change
Change divided by the original. Always divide by the old value.
e.g. 50 → 65: $\tfrac{15}{50} = 30\%$ increase
$\displaystyle \%\text{ change} = \frac{\text{new} - \text{old}}{\text{old}} \times 100\%$
4.43.3
Applying Percentage Change
Multiply the original by the multiplier to get the new value. Percent change is the change divided by the original value, not the new one.
e.g. A $60 jacket at $25\%$ off: $60 \times 0.75 = \$45$
$\displaystyle \text{new} = \text{original} \times (1 \pm r)$
$\displaystyle r = \frac{\text{new} - \text{old}}{\text{old}}$
4.43.4
Finding Original Values From Percentage Change
Undo the change by dividing the new value by the multiplier. Do not just subtract the percent from the new value.
e.g. After a $20\%$ raise the price is $72: $72 \div 1.2 = \$60$
$\displaystyle \text{original} = \frac{\text{new}}{1 \pm r}$
4.43.5
Applying Successive Percentage Changes
Multiply the multipliers, one per change. Never add the percents.
e.g. Up $20\%$ then down $20\%$ : $1.2 \times 0.8 = 0.96$ , a $4\%$ decrease
$\displaystyle \text{new} = \text{orig} \cdot (1 + r_1)(1 + r_2)$
$\displaystyle r < 0 \text{ for a decrease}$
4.43.6
Further Successive Percentage Changes
The same change applied $n$ times uses an exponent. To find an unknown change, set the product of the multipliers equal to the overall multiplier and solve.
e.g. Up $25\%$ , then back to the start: $1.25m = 1 \Rightarrow m = 0.8$ , a $20\%$ decrease
$\displaystyle \text{new} = \text{orig} \cdot (1 + r)^n$
4.43.7
Multistep Percentage Problems
Work one step at a time and track which whole each percent refers to. A percent of a percent multiplies.
e.g. 200 students, $40\%$ juniors, $25\%$ of juniors take chemistry: $0.25 \times 0.4 \times 200 = 20$
$\displaystyle \tfrac{p}{100} \cdot \tfrac{q}{100} \cdot N$

4.44 Unit Conversions

A conversion factor equals 1, so multiplying by it changes the units without changing the amount. The struck-through units cancel.

4.44.1
Unit Conversions to Smaller Units of Length
Going to a smaller unit gives a bigger number, so multiply by the conversion factor.
e.g. $3.5$ ft $= 3.5 \times 12 = 42$ in
$\displaystyle 1\text{ ft} = 12\text{ in},\; 1\text{ yd} = 3\text{ ft}$
$\displaystyle 1\text{ mi} = 5280\text{ ft}$
$\displaystyle 1\text{ m} = 100\text{ cm} = 1000\text{ mm}$
$\displaystyle 1\text{ km} = 1000\text{ m}$
4.44.2
Unit Conversions to Larger Units of Length
Going to a larger unit gives a smaller number, so divide by the conversion factor.
e.g. $90$ in $= 90 \div 12 = 7.5$ ft
$\displaystyle 1\text{ ft} = 12\text{ in},\; 1\text{ yd} = 3\text{ ft}$
$\displaystyle 1\text{ km} = 1000\text{ m}$
4.44.3
Unit Conversions Between Units of Mass and Volume
Use the same rule as length: multiply to go to a smaller unit, divide to go to a larger one.
e.g. $2.4$ kg $= 2400$ g. $3$ gal $= 12$ qt
$\displaystyle 1\text{ kg} = 1000\text{ g},\; 1\text{ g} = 1000\text{ mg}$
$\displaystyle 1\text{ lb} = 16\text{ oz},\; 1\text{ ton} = 2000\text{ lb}$
$\displaystyle 1\text{ L} = 1000\text{ mL}$
$\displaystyle 1\text{ gal} = 4\text{ qt},\; 1\text{ qt} = 2\text{ pt}$
$\displaystyle 1\text{ pt} = 2\text{ cups},\; 1\text{ cup} = 8\text{ fl oz}$
4.44.4
Two-Step Unit Conversions
Multiply by conversion fractions arranged so the unwanted units cancel.
$\displaystyle \frac{60\text{ mi}}{1\text{ h}} \cdot \frac{5280\text{ ft}}{1\text{ mi}} \cdot \frac{1\text{ h}}{3600\text{ s}} = 88\ \frac{\text{ft}}{\text{s}}$
4.44.5
Unit Conversions Between Mixed Units
For a compound unit like miles per hour, convert each unit separately. Multiply by fractions equal to 1, arranged so the old units cancel.
e.g. $60$ mi/h $= \tfrac{60 \times 5280}{3600}$ ft/s $= 88$ ft/s
$\displaystyle \frac{\text{mi}}{\text{h}} \cdot \frac{\text{ft}}{\text{mi}} \cdot \frac{\text{h}}{\text{s}} = \frac{\text{ft}}{\text{s}}$
4.44.6
Converting Between Mixed Units Containing Squares
A squared unit needs the conversion factor squared. Convert each unit in a compound rate separately.
e.g. $5\text{ m/s}^2 = 5 \times 60^2 = 18{,}000\text{ m/min}^2$
$\displaystyle 1\text{ ft} = 12\text{ in}$
$\displaystyle \Rightarrow\; 1\text{ ft}^2 = 144\text{ in}^2$

4.45 Ratios, Rates and Proportional Relationships

Tape diagram. A ratio is a count of equal shares; find one share and everything follows.

4.45.1
Ratios
A ratio $a:b$ compares two quantities in order. Simplify by dividing both by their GCF. A part-to-part ratio is different from a part-to-whole ratio.
e.g. 12 boys, 18 girls: boys to girls $= 2:3$ , boys to total $= 2:5$
$\displaystyle a : b = \frac{a}{b}$
4.45.2
Applying Ratios
For a ratio $a:b$ , write the quantities as $ak$ and $bk$ . If you know the total, each share is its part of $a + b$ .
e.g. Split $40 in the ratio $3:5$ : $\tfrac38 \times 40 = 15$ and $\tfrac58 \times 40 = 25$
$\displaystyle \text{share of } a = \frac{a}{a + b} \times \text{total}$
4.45.3
Working With Ratios in Various Settings
To combine two ratios that share a quantity, scale them so the shared quantity matches. Ratios stay the same when both terms are multiplied by the same number.
e.g. Recipe uses flour to sugar $3:1$ . With 2 cups sugar, use $3 \times 2 = 6$ cups flour
$\displaystyle a:b = 2:3,\; b:c = 4:5$
$\displaystyle \Rightarrow\; a:b:c = 8:12:15$
4.45.4
Rate Problems
A rate is a quantity per one unit of another quantity. Find the unit rate, then multiply.
e.g. 270 words in 3 min is 90 words/min, so 7 min gives $630$ words
$\displaystyle \text{rate} = \frac{\text{amount}}{\text{time}}$
$\displaystyle \text{amount} = \text{rate} \times \text{time}$
4.45.5
Multistep Rate Problems
Chain the rates so the units cancel, one step at a time. Write units on every number to check the setup.
e.g. 150 mi at 30 mi/gal, gas at $4/gal: $\tfrac{150}{30} = 5$ gal, $5 \times 4 = \$20$
4.45.6
Speed Problems
Distance equals speed times time. Average speed is total distance divided by total time, not the average of the speeds.
e.g. 60 mi at 30 mph (2 h) then 60 mi at 60 mph (1 h): $\tfrac{120}{3} = 40$ mph
$\displaystyle d = rt$
$\displaystyle \text{avg speed} = \frac{\text{total distance}}{\text{total time}}$
4.45.7
Algebraic Expressions With Rates and Ratios
Build the expression the same way you would with numbers, then replace the numbers with the variables. Watch for unit changes such as minutes to hours.
e.g. A machine makes $m$ parts per hour. In $t$ minutes it makes $\tfrac{mt}{60}$ parts
4.45.8
Solving Equations Describing Proportional Relationships
$y$ is proportional to $x$ when $y = kx$ for a constant $k$ . Find $k$ from one known pair, then use it. To solve a proportion, cross-multiply.
e.g. $y = 12$ when $x = 4$ , so $k = 3$ . When $x = 7$ , $y = 21$
$\displaystyle y = kx,\quad k = \frac{y}{x}$
$\displaystyle \frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc$
4.45.9
Calculating Density
Density is mass per unit of volume. Population density is people per unit of area. Rearrange to find mass or volume.
e.g. 54 g in 20 cm³: $\tfrac{54}{20} = 2.7$ g/cm³
$\displaystyle \text{density} = \frac{\text{mass}}{\text{volume}}$
$\displaystyle \text{mass} = \text{density} \times \text{volume}$
4.45.10
Descriptions of Physical Phenomena
Translate the description into a rate equation. The amount equals the starting value plus the rate times time, with a negative rate for a decrease. The units of the answer come from the units of the inputs.
e.g. A 120 gal tank drains at 3 gal/min: $A = 120 - 3t$ , empty at $t = 40$ min
$\displaystyle \text{amount} = \text{start} + \text{rate} \times t$

No topic matches that search. Try a single word, like “slope” or “area.”

Topic numbers match the course outline, so a topic here lines up with the same number in the lesson list.

1.1 Linear Expressions and Equations in One Variable

Evaluating Expressions

Finding Equivalent Equations

Solving One-Step Addition Equations

Solving One-Step Multiplication Equations

Solving Multistep Linear Equations

Solving Equations by Combining Terms

Solving Equations With Fractional Coefficients

Solving Linear Equations Containing Binomials

Solving Linear Equations Containing Binomials by Factoring

Solving Linear Equations With Binomials on Both Sides

Manipulating Linear Equations Containing Binomials

1.2 Word Problems and Modeling

Numerical Word Problems

Multistep Numerical Word Problems

Constructing Models With One-Variable Linear Equations

Modeling and Solving Problems Using Linear Equations

Linear Models Presented in Unsimplified Form

Modeling Geometry Problems Using Linear Equations

1.3 Two-Variable Linear Equations

Solving Two-Variable Linear Models

Constructing Models With Two-Variable Linear Equations

Modeling Totals Using Two-Variable Linear Equations

Interpreting Two-Variable Linear Equations

Interpreting Coefficients of Two-Variable Linear Equations

1.4 Linear Functions

Evaluating Linear Functions

Finding a Function Input for a Given Output

Linear Functions Containing Unknown Parameters

Working With Function Composition

1.5 Functions in Tabular Form

Identifying Tables Corresponding to Linear Functions

Identifying Linear Functions Corresponding to Tables

Linear Functions as Tables in Applications

Analyzing Functions in Tabular Form

Function Tables Containing Unknown Parameters

1.6 Graphs of Linear Equations

Interpreting Coefficients of Lines in Slope-Intercept Form

Finding Equations of Lines Using Slopes and Points

Graphs of Linear Equations in Slope-Intercept Form

Finding Intercepts of Lines in Slope-Intercept Form

Constructing and Analyzing Slope-Intercept Equations

Properties of Lines Given in Standard or General Form

Analyzing Linear Equations in General Form

Analyzing Linear Equations of Shifted Lines

Constructing Linear Functions Using Slope-Intercept Form

1.7 Parallel and Perpendicular Lines

Parallel Lines

Slopes of Perpendicular Lines

Perpendicular Lines

1.8 Modeling With Linear Functions

Modeling With Linear Functions

Constructing Models Using Linear Functions

Further Constructing Models Using Linear Functions

Interpreting Coefficients of Linear Functions in Context

Interpreting Graphs of Linear Functions

Modeling Using Slope-Intercept Form

1.9 Systems of Two-Variable Linear Equations

Solving Simple Systems of Linear Equations

Identifying Structure in Systems of Linear Equations

Solving Systems of Equations Using Addition or Subtraction

Solving Systems of Equations Using Elimination

Modeling Word Problems With Systems of Linear Equations

Solving Problems With Systems of Linear Equations

Solving Problems With Systems of Linear Equations in Standard Form

The Intersection of Two Lines

The Relationship Between Systems of Linear Equations and Graphs

1.10 Linear Equations With Infinitely Many or No Solutions

Determining the Number of Solutions to a Linear Equation

Linear Equations With Infinitely Many Solutions

Linear Equations With No Solutions

Determining the Number of Solutions to a Linear System

Linear Systems With Infinitely Many Solutions

Parametrizing Systems of Linear Equations

Linear Systems With No Solution

1.11 Linear Inequalities One Variable

Constructing Models With One-Variable Linear Inequalities

Modeling and Solving With One-Variable Linear Inequalities

Modeling and Solving Percentage Problems With One-Variable Linear Inequalities

Modeling and Solving With One-Variable Linear Inequalities: Advanced Cases