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1. Function notation: evaluate and read f(x)

Discover

$f(x) = x + 5$ is a rule: it takes a number and adds 5.

What does $f(2)$ mean?

The output of $f$ when the input is 2
$f$ times 2
The input that gives the output 2

Key idea

A function gives exactly one output for each input. We write the output as $f(x)$ .

f(a)

is the output of

f

for the input

a

. To find it, replace every

x

in the rule by

(a)

, with parentheses.

On a graph, $f(a)$ is the height of the graph above $x = a$ .

Worked example

$f(x) = 3x^2 - x$ . Find $f(-2)$ .

f(-2) = 3(-2)^2 - (-2)

Replace each x by (−2), with the parentheses.

= 3 \cdot 4 + 2

(−2)² = 4, and subtracting −2 adds 2.

= 14

Multiply, then add.

Your turn, guided

Your turn: $f(x) = 2x^2 + 1$ . Fill in the missing line.

f(-3) = 2(-3)^2 + 1

Replace x by (−3).

f(-3) = 19

(−3)² = 9, then 2 · 9 + 1.

Practice ★

$f(x) = 2x^2 + 3$ . Find $f(-3)$ .
$f(x) = 4x^2 + 8$ . Find $f(-2)$ .
$f(x) = 4x^2 + 2$ . Find $f(-5)$ .
$f(x) = 5x^2 + 5$ . Find $f(-4)$ .
$f(x) = 3x^2 + 3$ . Find $f(-4)$ .

Watch out!

Practice ★★

$f(x) = x^2 + 3$ . Find $f(2 + 3)$ .
$f(x) = x^2 + 2$ . Find $f(4 + 4)$ .
$f(x) = x^2 + 7$ . Find $f(4 + 1)$ .
$f(x) = x^2 + 3$ . Find $f(3 + 3)$ .

Here is the graph of $y = f(x)$ . Tap the point of the graph that shows $f(2)$ .
Here is the graph of $y = f(x)$ . Tap the point of the graph that shows $f(2)$ .

Exit ticket

Three questions to check you have got it.

$f(x) = x^2 + 3$ . Find $f(2 + 3)$ .
$f(x) = x^2 + 2$ . Find $f(4 + 4)$ .
$f(x) = x^2 + 7$ . Find $f(4 + 1)$ .

Summary card

Function notation

$f(a)$ is the output for the input $a$ .
Replace every $x$ by $(a)$ , with parentheses: $f(-2)$ for $x^2$ is $(-2)^2 = 4$ .
On a graph: the height above $x = a$ .

Watch out: $f(a + b)$ is not $f(a) + f(b)$ .

2. Piecewise-defined functions and absolute value

Warm-up

Three quick questions from earlier lessons.

Discover

A parking lot charges 5 dollars for up to 2 hours, and 9 dollars for a longer stay. What does a stay of exactly 2 hours cost?

5 dollars
9 dollars

Key idea

A piecewise function uses a different formula on different parts of its domain.

To find

f(a)

: first find which condition

a

satisfies, then use that piece's formula. Check the boundary carefully (

<

\le

The absolute value is piecewise: $|x| = x$ if $x \ge 0$ , and $|x| = -x$ if $x < 0$ .

Worked example

$f(x) = \begin{cases} x^2 & \text{if } x < 1 \\ 2x + 3 & \text{if } x \ge 1 \end{cases}$ . Find $f(1)$ .

Use the second piece,

2x + 3

Check the conditions first: 1 is not less than 1, but 1 ≥ 1.

f(1) = 2 \cdot 1 + 3 = 5

Replace x by 1.

Your turn, guided

Your turn, with the same $f$ : find $f(-2)$ .

Use the first piece,

x^2

−2 < 1.

f(-2) = 4

(−2)² = 4.

Practice ★

$f(x) = \begin{cases} x + 4 & \text{if } x < 8 \\ 4x & \text{if } x \ge 8 \end{cases}$ . Find $f(0)$ .
$f(x) = \begin{cases} x + 8 & \text{if } x < 7 \\ 4x & \text{if } x \ge 7 \end{cases}$ . Find $f(2)$ .
$f(x) = \begin{cases} x + 7 & \text{if } x < 3 \\ 3x & \text{if } x \ge 3 \end{cases}$ . Find $f(1)$ .
$f(x) = \begin{cases} x + 6 & \text{if } x < 6 \\ 4x & \text{if } x \ge 6 \end{cases}$ . Find $f(1)$ .

Watch out!

Practice ★★

$f(x) = \begin{cases} x + 3 & \text{if } x < 5 \\ 3x & \text{if } x \ge 5 \end{cases}$ . Find $f(5)$ .
$f(x) = \begin{cases} x + 4 & \text{if } x < 3 \\ 2x & \text{if } x \ge 3 \end{cases}$ . Find $f(3)$ .
$f(x) = \begin{cases} x + 1 & \text{if } x < 1 \\ 5x & \text{if } x \ge 1 \end{cases}$ . Find $f(1)$ .
$f(x) = \begin{cases} x + 2 & \text{if } x < 5 \\ 2x & \text{if } x \ge 5 \end{cases}$ . Find $f(5)$ .

For $x < 7$ , which expression is equal to $|x - 7|$ ?
1. $7 - x$
2. $x - 7$
3. $-x - 7$
For $x < 5$ , which expression is equal to $|x - 5|$ ?
1. $5 - x$
2. $x - 5$
3. $-x - 5$

Exit ticket

Three questions to check you have got it.

$f(x) = \begin{cases} x + 3 & \text{if } x < 5 \\ 3x & \text{if } x \ge 5 \end{cases}$ . Find $f(5)$ .
$f(x) = \begin{cases} x + 4 & \text{if } x < 3 \\ 2x & \text{if } x \ge 3 \end{cases}$ . Find $f(3)$ .
$f(x) = \begin{cases} x + 1 & \text{if } x < 1 \\ 5x & \text{if } x \ge 1 \end{cases}$ . Find $f(1)$ .

Summary card

Piecewise functions

Find the piece whose condition the input satisfies.
Use that piece's formula.
$|x| = x$ for $x \ge 0$ , and $-x$ for $x < 0$ .

Watch out: at a boundary, read $<$ and $\ge$ carefully.

3. Transformations: shifts and reflections

Warm-up

Three quick questions from earlier lessons.

Discover

Move the sliders and watch the graph of $y = (x - h)^2 + k$ .

The graph of y = (x − h)² + k, a parabola whose vertex is at (h, k) and moves with the sliders.

When $h$ goes from 0 to 2, which way does the graph move?

To the right
To the left

Key idea

Changing the rule moves or flips the graph without changing its shape.

y = f(x - h) + k

moves the graph of

f

h

units to the right and

k

units up.

y = -f(x)

reflects it across the

x

-axis;

y = f(-x)

reflects it across the

y

-axis.

A change inside $f(\dots)$ acts on $x$ (horizontally); a change outside acts on $y$ (vertically).

Worked example

Describe $y = (x + 1)^2 - 3$ compared with $y = x^2$ .

1 unit to the left

x + 1 = x − (−1): h = −1.

3 units down

− 3 outside: k = −3.

Vertex:

(-1, -3)

The vertex (0, 0) moves with the graph.

Your turn, guided

Your turn: $y = (x - 4)^2 + 2$ .

4 units to the right

x − 4: h = 4.

2 units up

+ 2 outside.

The vertex is at

x = 4

The vertex moves from x = 0 to x = …

Practice ★

The graph of $y = x^2$ is moved 1 units to the right and 6 units down. Write the equation of the new graph.
The graph of $y = x^2$ is moved 3 units to the right and 1 units down. Write the equation of the new graph.
The graph of $y = x^2$ is moved 4 units to the right and 6 units down. Write the equation of the new graph.
The graph of $y = x^2$ is moved 7 units to the right and 1 units down. Write the equation of the new graph.

Watch out!

Practice ★★

Draw the graph of $y = (x - 1)^2 - 3$ : place the vertex and one more point.
Draw the graph of $y = (x - 1)^2 - 1$ : place the vertex and one more point.

Match each equation with what it does to the graph of $y = f(x)$ .
1. $y = f(x) + 3$
2. $y = f(x - 3)$
3. $y = -f(x)$
4. $y = f(-x)$
1. moves it up 3 units
2. moves it 3 units to the right
3. reflects it across the x-axis
4. reflects it across the y-axis

Exit ticket

Three questions to check you have got it.

The graph of $y = x^2$ is moved 1 units to the right and 6 units down. Write the equation of the new graph.
The graph of $y = x^2$ is moved 3 units to the right and 1 units down. Write the equation of the new graph.
The graph of $y = x^2$ is moved 4 units to the right and 6 units down. Write the equation of the new graph.

Summary card

Shifts and reflections

$f(x - h)$ : $h$ units right. $f(x) + k$ : $k$ units up.
$-f(x)$ : flip across the $x$ -axis. $f(-x)$ : flip across the $y$ -axis.
Inside $f(\dots)$ acts on $x$ ; outside acts on $y$ .

Watch out: $f(x + 3)$ moves the graph left.

4. Stretches and compressions; even and odd functions

Warm-up

Three quick questions from earlier lessons.

Discover

Move the slider and watch the graph of $y = (bx)^2$ .

The graph of y = (bx)², a parabola that gets narrower as b grows.

When $b$ goes from 1 to 2, the graph becomes…

narrower
wider

Key idea

Multiplying stretches or squeezes a graph. Symmetry has a name.

For

a > 1

y = a f(x)

stretches the graph vertically by

a

. For

b > 1

y = f(bx)

compresses it horizontally by

b

f

is even if

f(-x) = f(x)

(symmetric about the

y

-axis), odd if

f(-x) = -f(x)

(symmetric about the origin).

Worked example

Is $f(x) = x^3 - 2x$ even, odd, or neither?

f(-x) = (-x)^3 - 2(-x)

Replace x by (−x).

= -x^3 + 2x

(−x)³ = −x³ and −2(−x) = 2x.

= -(x^3 - 2x) = -f(x)

: odd

Factor out −1: this is −f(x).

Your turn, guided

Your turn: $f(x) = x^4 + 3$ . Check one value.

f(-1) = (-1)^4 + 3

Replace x by −1.

f(-1) = 4

(−1)⁴ = 1. It equals f(1): a sign that f is even.

Practice ★

How does the graph of $y = f(4x)$ compare with the graph of $y = f(x)$ ?
1. It is squeezed toward the y-axis: horizontal compression by a factor of 4
2. It is pulled away from the y-axis: horizontal stretch by a factor of 4
3. It is stretched vertically by a factor of 4
How does the graph of $y = f(5x)$ compare with the graph of $y = f(x)$ ?
1. It is squeezed toward the y-axis: horizontal compression by a factor of 5
2. It is pulled away from the y-axis: horizontal stretch by a factor of 5
3. It is stretched vertically by a factor of 5
How does the graph of $y = f(3x)$ compare with the graph of $y = f(x)$ ?
1. It is squeezed toward the y-axis: horizontal compression by a factor of 3
2. It is pulled away from the y-axis: horizontal stretch by a factor of 3
3. It is stretched vertically by a factor of 3

Watch out!

Practice ★★

$f(x) = 8x^3 - 9x$ . Is $f$ even, odd, or neither?
1. Even
2. Odd
3. Neither
$f(x) = 9x^3 - 2x$ . Is $f$ even, odd, or neither?
1. Even
2. Odd
3. Neither

$f(x) = 3x^4 - 8x^2 + 7$ . Is $f$ even, odd, or neither?
1. Even
2. Odd
3. Neither
$f(x) = 5x^4 - 1x^2 + 9$ . Is $f$ even, odd, or neither?
1. Even
2. Odd
3. Neither

Exit ticket

Three questions to check you have got it.

$f(x) = 8x^3 - 9x$ . Is $f$ even, odd, or neither?
1. Even
2. Odd
3. Neither
$f(x) = 9x^3 - 2x$ . Is $f$ even, odd, or neither?
1. Even
2. Odd
3. Neither
$f(x) = 7x^3 - 1x$ . Is $f$ even, odd, or neither?
1. Even
2. Odd
3. Neither

Summary card

Stretches, even and odd

$a f(x)$ : vertical stretch by $a$ . $f(bx)$ : horizontal compression by $b$ .
Even: $f(-x) = f(x)$ , mirror image in the $y$ -axis.
Odd: $f(-x) = -f(x)$ , symmetric about the origin.

Watch out: compute $f(-x)$ ; the signs of the coefficients do not decide.

5. Composition of functions

Warm-up

Three quick questions from earlier lessons.

Discover

$g$ doubles a number: $g(x) = 2x$ . $f$ adds 1: $f(x) = x + 1$ . Start with 3: double it, then add 1. You have computed $f(g(3))$ . What is it?

Key idea

Composing means using the output of one function as the input of another.

f(g(x))

: first apply

g

x

, then apply

f

to the result. Work from the inside out. In general

f(g(x)) \ne g(f(x))

Worked example

$f(x) = x^2$ and $g(x) = x - 3$ . Find $f(g(5))$ , then $f(g(x))$ .

g(5) = 5 - 3 = 2

The inside first.

f(g(5)) = f(2) = 4

Then f of that result.

f(g(x)) = (x - 3)^2

With x: put g(x) = x − 3 in place of x in f.

Your turn, guided

Your turn: $f(x) = 2x + 1$ and $g(x) = x^2$ . Find $f(g(3))$ .

g(3) = 9

The inside first.

f(g(3)) = f(9) = 19

Then 2 · 9 + 1.

Practice ★

$f(x) = 2x + 2$ and $g(x) = x^2 - 2$ . Find $f(g(3))$ .
$f(x) = 3x + 3$ and $g(x) = x^2 - 4$ . Find $f(g(3))$ .
$f(x) = 4x + 3$ and $g(x) = x^2 - 1$ . Find $f(g(2))$ .
$f(x) = 3x + 1$ and $g(x) = x^2 - 5$ . Find $f(g(3))$ .

Watch out!

Practice ★★

$f(x) = 6x + 6$ and $g(x) = x + 5$ . Write $f(g(x))$ .
$f(x) = 5x + 7$ and $g(x) = x + 2$ . Write $f(g(x))$ .
$f(x) = 6x + 8$ and $g(x) = x + 3$ . Write $f(g(x))$ .

$g(1) = 5$ , $g(5) = 7$ , $f(1) = 7$ and $f(5) = 1$ . Find $f(g(1))$ .
$g(2) = 6$ , $g(6) = 7$ , $f(2) = 7$ and $f(6) = 2$ . Find $f(g(2))$ .

Exit ticket

Three questions to check you have got it.

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

Summary card

Composition

$f(g(x))$ : apply $g$ first, then $f$ .
Work from the inside out.
The order matters: $f(g(x))$ and $g(f(x))$ usually differ.

Watch out: $f(g(x))$ is not $f(x) \cdot g(x)$ .

6. Inverse functions

Warm-up

Three quick questions from earlier lessons.

Discover

$f(x) = x + 7$ adds 7. Which function undoes it, and gives back the number you started with?

$x - 7$
$\dfrac{1}{x + 7}$
$x + 7$

Key idea

The inverse function $f^{-1}$ undoes $f$ : it swaps inputs and outputs.

f(a) = b

, then

f^{-1}(b) = a

. To find

f^{-1}(x)

: write

y = f(x)

, solve for

x

, then rename. The graph of

f^{-1}

is the reflection of the graph of

f

across the line

y = x

$f^{-1}(x)$ is not $\dfrac{1}{f(x)}$ : the $-1$ is not an exponent here.

Worked example

$f(x) = 3x - 5$ . Find $f^{-1}(x)$ .

y = 3x - 5

Name the output y.

y + 5 = 3x

f subtracts 5 last, so undo that first: add 5.

x = \dfrac{y + 5}{3}

Then undo × 3: divide by 3.

f^{-1}(x) = \dfrac{x + 5}{3}

Rename the input x.

Your turn, guided

Your turn: $f(x) = 2x + 6$ .

y = 2x + 6

Name the output y.

y - 6 = 2x

Undo + 6 first.

f^{-1}(x) = \dfrac{x - 6}{2}

Divide by 2, then rename.

Practice ★

$f(x) = 7x + 9$ . Find $f^{-1}(x)$ .
$f(x) = 2x + 2$ . Find $f^{-1}(x)$ .
$f(x) = 7x + 5$ . Find $f^{-1}(x)$ .
$f(x) = 2x + 9$ . Find $f^{-1}(x)$ .

Watch out!

Practice ★★

$f(x) = 2x + 6$ . Find $f^{-1}(10)$ .
$f(x) = 2x + 9$ . Find $f^{-1}(11)$ .
$f(x) = 3x + 1$ . Find $f^{-1}(10)$ .

The point $(2, 7)$ is on the graph of $f$ . Tap the point that must be on the graph of $f^{-1}$ .
The point $(5, 7)$ is on the graph of $f$ . Tap the point that must be on the graph of $f^{-1}$ .

Exit ticket

Three questions to check you have got it.

$f(x) = 7x + 9$ . Find $f^{-1}(x)$ .
$f(x) = 2x + 2$ . Find $f^{-1}(x)$ .
$f(x) = 7x + 5$ . Find $f^{-1}(x)$ .

Summary card

Inverse functions

Write $y = f(x)$ .
Solve for $x$ , undoing the last operation first.
Rename: $f^{-1}(x)$ . Check that $f(f^{-1}(x)) = x$ .

Watch out: $f^{-1}(x)$ is not $\dfrac{1}{f(x)}$ .

Answers: Transformations of functions

1. Function notation: evaluate and read f(x)

$21$
$24$
$102$
$85$
$51$

$28$
$66$
$32$
$39$

$(2, -1)$
$(2, 3)$

$28$
$66$
$32$

2. Piecewise-defined functions and absolute value

$4$
$10$
$8$
$7$

$15$
$6$
$5$
$10$

$7 - x$
$5 - x$

$15$
$6$
$5$

3. Transformations: shifts and reflections

$\left(x - 1\right)^{2} - 6$
$\left(x - 3\right)^{2} - 1$
$\left(x - 4\right)^{2} - 6$
$\left(x - 7\right)^{2} - 1$

$y = \left(x - 1\right)^{2} - 3$
$y = \left(x - 1\right)^{2} - 1$

$y = f(x) + 3$ → moves it up 3 units · $y = f(x - 3)$ → moves it 3 units to the right · $y = -f(x)$ → reflects it across the x-axis · $y = f(-x)$ → reflects it across the y-axis

$\left(x - 1\right)^{2} - 6$
$\left(x - 3\right)^{2} - 1$
$\left(x - 4\right)^{2} - 6$

4. Stretches and compressions; even and odd functions

It is squeezed toward the y-axis: horizontal compression by a factor of 4
It is squeezed toward the y-axis: horizontal compression by a factor of 5
It is squeezed toward the y-axis: horizontal compression by a factor of 3

Even
Even

5. Composition of functions

$16$
$18$
$15$
$13$

$6 x + 36$
$5 x + 17$
$6 x + 26$

$16$
$18$
$15$

6. Inverse functions

$\frac{x}{7} - \frac{9}{7}$
$\frac{x}{2} - 1$
$\frac{x}{7} - \frac{5}{7}$
$\frac{x}{2} - \frac{9}{2}$

$(7, 2)$
$(7, 5)$

$\frac{x}{7} - \frac{9}{7}$
$\frac{x}{2} - 1$
$\frac{x}{7} - \frac{5}{7}$