Functions and Their Graphs — AP® Calculus AB

Part 1 · Hook

Why this matters

Every question in calculus is a question about a function: how fast a tank fills, where a ball is at time

t

, how much a quantity has piled up. Before calculus asks how a function changes, you need to read one fluently, from its formula, its graph, a table or a sentence. This page is a fast, friendly tune-up of the function skills the course leans on most.

Part 2 · See it

See it first

Graph of a piecewise function f from x = -4 to x = 5: a parabola piece from the filled point (-4, 1) down to (-2, -1) and up toward an open circle at (1, 3.5), then a line from the filled point (1, 5) down to the filled point (5, 1). Notes beside it: f(1) = 5 is the filled dot; the open circle is not on the graph; the range is from -1 to 5; the line y = 2 meets the graph at x = 0 and x = 4. — Reading a graph: $f(1)$ is the height of the filled dot, an open circle is a point left out, and the domain and range come from the graph's left-right and up-down extent. LevlPrep original diagram.

Part 3 · Step by step

How it works, step by step

A function assigns exactly one output to each input $f(3)$ names one number: the height of the graph above $x = 3$
Some inputs break a formula (a zero denominator, the square root of a negative)those inputs are left out of the domain
Feeding one function's output into another builds a composite function $f(g(x))$ means: do $g$ first, then $f$
Replacing $x$ by $x - h$ and adding $k$ to the outputmoves the whole graph $h$ units right and $k$ units up

Part 4 · Key ideas

Key ideas

A function gives one output for each input. Substitute with parentheses: if $f(x) = x^2 - 3x$ , then $f(a + 1)$ $= (a + 1)^2 - 3(a + 1)$ .
The domain is every input the rule accepts; the range is every output it produces. Rule out zero denominators and square roots of negatives.
On a graph, a filled dot is on the graph and an open circle is not. $f(a)$ is the height of the graph at $x = a$ .
A line through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$ (point-slope form). Slope is rise over run.
$f(g(x))$ means $g$ first, then $f$ . Order matters: usually $f(g(x)) \ne g(f(x))$ .
A rational function has a vertical asymptote where the denominator is 0 and the numerator is not.

Part 5 · Misconception

A common mistake

The wrong idea: $f(g(x))$ means $f(x)$ times $g(x)$ .

What actually happens: $f(g(x))$ is a composition: the output of $g$ becomes the input of $f$ . With $f(x) = x^2$ and $g(x) = x + 1$ , $f(g(x)) = (x + 1)^2$ , while $f(x)g(x) = x^2(x + 1)$ .

Part 6 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $g$

The function

g

is defined on

-3 \le x \le 4

. Its graph is a line segment from

(-3, 2)

toward an open circle at

(1, -2)

, then a curved piece from the filled point

(1, 2)

through

(3, 0)

to

(4, 0.5)

.

1. What is $g(1)$ ?

$2$
$-2$
$0$
$g(1)$ is not defined

Show the answer

At $x = 1$ the filled dot is at height 2, so $g(1) = 2$ . The open circle at $(1, -2)$ marks a point that is not on the graph.

Correct: $2$ : The filled dot at $x = 1$ sits at height 2.
$-2$ : That height belongs to the open circle, which is not part of the graph.
$0$ : The curved piece reaches 0 at $x = 3$ , not at $x = 1$ .
$g(1)$ is not defined: There is a filled dot at $x = 1$ , so $g(1)$ is defined.

2. What is the range of $g$ ?

$-2 < y \le 2$
$-2 \le y \le 2$
$0 \le y \le 2$
$-3 \le y \le 4$

Show the answer

The line takes every height from 2 down toward $-2$ , but $-2$ itself is the open circle, so it is left out. The curve's heights, from 0 to 2, are already inside. Range: $-2 < y \le 2$ .

Correct: $-2 < y \le 2$ : The line covers $(-2, 2]$ ; the curve's heights $[0, 2]$ add nothing new.
$-2 \le y \le 2$ : The height $-2$ occurs only at the open circle, which is not on the graph.
$0 \le y \le 2$ : The line piece also gives every negative height down toward $-2$ .
$-3 \le y \le 4$ : Those are the domain's endpoints; the range is about outputs (heights).

3. What is $g(g(4))$ ?

$-1.5$
$0.5$
$0.25$
$3.125$

Show the answer

Work from the inside out. $g(4) = 0.5$ . Then $g(0.5)$ uses the line, because $0.5 < 1$ : $g(0.5) = -0.5 - 1 = -1.5$ .

Correct: $-1.5$ : $g(4) = 0.5$ , and $g(0.5) = -0.5 - 1 = -1.5$ on the line.
$0.5$ : That is only the inside step, $g(4)$ . Feed 0.5 back into $g$ .
$0.25$ : That is $g(4) \cdot g(4)$ . A composition feeds one output into the function again; it does not multiply.
$3.125$ : That uses the curved piece's formula at $x = 0.5$ , but the curve only covers $1 \le x \le 4$ .

4. Let $f(x) = x^2 + 1$ and $g(x) = 2x - 3$ . Find $f(g(x))$ .

Type an expression for $f(g(x)) =$ .

Show the answer

Replace $x$ in $f$ by $(2x - 3)$ : $f(g(x)) = (2x - 3)^2 + 1$ , which expands to $4x^2 - 12x + 10$ . Either form is right.

Answer: $f(g(x)) =$ $(2x - 3)^{2} + 1$

5. Find an equation of the line through $(-1, 5)$ and $(3, -3)$ . Write $y$ in terms of $x$ .

Type an expression for $y =$ .

Show the answer

Slope: $m$ $= \frac{-3 - 5}{3 - (-1)}$ $= -2$ . Point-slope form with $(-1, 5)$ : $y - 5 = -2(x + 1)$ , so $y = -2x + 3$ . Check with $(3, -3)$ : $-6 + 3 = -3$ .

Answer: $y =$ $-2 x + 3$

6. The graph of $y = |x|$ is moved 3 units left and 2 units down. Which equation describes the new graph?

$y = |x + 3| - 2$
$y = |x - 3| - 2$
$y = |x + 3| + 2$
$y = |x - 2| + 3$

Show the answer

Replacing $x$ by $x + 3$ moves a graph 3 units left; subtracting 2 outside moves it 2 units down. The point of the V moves from $(0, 0)$ to $(-3, -2)$ .

Correct: $y = |x + 3| - 2$ : Left 3 is $x + 3$ inside; down 2 is $-2$ outside.
$y = |x - 3| - 2$ : $x - 3$ inside moves the graph right, not left.
$y = |x + 3| + 2$ : $+2$ outside moves the graph up, not down.
$y = |x - 2| + 3$ : This mixes up which number goes inside and the directions: it moves the graph right 2 and up 3.

7. Let $f(x)$ $= \begin{cases} x^2 & x < 2 \\ 3x - 1 & x \ge 2 \end{cases}$ . What is $f(2) + f(-1)$ ?

$6$
$5$
$1$
$4$

Show the answer

$x = 2$ satisfies $x \ge 2$ , so $f(2) = 3(2) - 1 = 5$ . $x = -1$ satisfies $x < 2$ , so $f(-1) = (-1)^2 = 1$ . The sum is 6.

Correct: $6$ : $f(2) = 5$ from the second piece and $f(-1) = 1$ from the first.
$5$ : That uses $x^2$ at $x = 2$ , giving $4 + 1$ . The boundary point 2 belongs to the piece with $x \ge 2$ .
$1$ : That uses $3x - 1$ at $x = -1$ , giving $5 + (-4)$ . Since $-1 < 2$ , use $x^2$ .
$4$ : $(-1)^2 = 1$ , not $-1$ : the parentheses keep the sign inside the square.

Part 7 · Summary

Summary

A function sends each input to one output. Read it from a formula (substitute with parentheses), a graph (heights, filled and open dots) or a table. Know the domain, write lines in point-slope form, compose functions inside first, move graphs with

f(x - h) + k

, and find vertical asymptotes where a denominator is 0.

Part 8 · Up next

What comes next

Part 9 · Connections

Connections

Next: Exponent Rules and Rewriting Powers

Why this matters

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of gg

Summary

What comes next

Connections

The graph of $g$