Unit 1 · Topic 1.12 · about 30 minutes · 7 parts · 11 scored checks, 2 of them multiple choice

Transformations of Functions

Built from College Board's course framework · Every answer key independently checked

The parent graph y = x squared, dashed, and its transformation g(x) = 2(x - 3) squared - 4, solid. The vertex moves from (0, 0) to (3, -4), and the solid graph is twice as tall at each step away from its vertex.

Part 1 Think first

Which way does it move?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

The graph of $f$ is known. A new function is $g (x) = f (x + 3)$ . Compared with the graph of $f$ , the graph of $g$ is moved

In short. A transformation builds a new function from an old one by adding to or multiplying the input or the output, which moves, stretches or reflects the graph.

$f (x) + k$ translates the graph vertically by $k$ ; $f (x + h)$ translates it horizontally by $- h$ .
$a f (x)$ dilates vertically by a factor of $| a |$ , reflecting over the $x$ -axis if $a < 0$ ; $f (b x)$ dilates horizontally by a factor of $| \frac{1}{b} |$ , reflecting over the $y$ -axis if $b < 0$ .
Transformations combine, taking a preimage to an image, and the domain and range can change along the way.

Part 2 Learn it

Three short ideas, each checked as you go

2a. Translations: adding

An additive transformation adds a constant.

$g (x) = f (x) + k$ adds $k$ to every output, a vertical translation of the graph of $f$ by $k$ units: up if $k > 0$ , down if $k < 0$ .
$g (x) = f (x + h)$ adds $h$ to every input before $f$ acts, a horizontal translation by $- h$ units.

The second one is the trap. $g (x) = f (x + 4)$ reaches each output of $f$ when $x$ is 4 less than before, so the graph moves 4 units left. $f (x - 4)$ moves it 4 units right.

In the diagram, the parent $y = x^{2}$ has its vertex at $(0, 0)$ . Replacing $x$ with $x - 3$ and then subtracting 4 moves the vertex to $(3, - 4)$ .

Quick check · Match

No calculator. Tap a function, then how its graph compares with the graph of $f$ .

2b. Dilations and reflections: multiplying

A multiplicative transformation multiplies by a constant.

$g (x) = a f (x)$ , with $a \neq 0$ , multiplies every output by $a$ : a vertical dilation by a factor of $| a |$ . If $a < 0$ , it also reflects the graph over the $x$ -axis.
$g (x) = f (b x)$ , with $b \neq 0$ , multiplies every input by $b$ : a horizontal dilation by a factor of $| \frac{1}{b} |$ . If $b < 0$ , it also reflects the graph over the $y$ -axis.

Again the input change works the opposite way. $f (2 x)$ reaches $f (6)$ when $x = 3$ , so every point moves halfway to the $y$ -axis: the graph is squeezed horizontally by a factor of $\frac{1}{2}$ . A point $(6, 2)$ on $f$ becomes $(3, 2)$ on $f (2 x)$ , and $(6, 6)$ on $3 f (x)$ .

Quick check · Calculate

No calculator. The point $(6, 2)$ is on the graph of $f$ . The corresponding point on the graph of $g (x) = 3 f (2 x)$ is $(p, q)$ . What is $p$ ?

Your answer

2c. Combining, and what happens to domain and range

Transformations combine. In $g (x) = 2 (x - 3)^{2} - 4$ , the parent $y = x^{2}$ is translated right 3, dilated vertically by 2, and translated down 4. The set of points before the transformations is the preimage; the set after is the image. Each point of the preimage has a partner in the image: $(1, 1)$ goes to $(4, 2 \cdot 1 - 4) = (4, - 2)$ .

The domain and range can change. The range of $y = x^{2}$ is $y \geq 0$ . Multiplying by $- 2$ flips that to $y \leq 0$ , and adding 5 moves it to $y \leq 5$ . If $f$ is only defined for $0 \leq x \leq 4$ , then $f (x - 1)$ is defined for $1 \leq x \leq 5$ : horizontal changes move the domain, vertical changes move the range.

Quick check · Multiple choice

No calculator. Let $f (x) = x^{2}$ , whose range is $y \geq 0$ . What is the range of $g (x) = - 2 f (x) + 5$ ?

Part 3 Watch it done

Building a transformed function

Watch each step, then do the last one yourself.

Start from $f (x) = x^{2}$ . Write a function $g$ whose graph has its vertex at $(3, - 4)$ and is dilated vertically by a factor of 2. Then find $g (5)$ .

Step 1 · Move the vertex sideways
A vertex at $x = 3$ needs a horizontal translation by 3, which is $f (x - 3) = (x - 3)^{2}$ .
Step 2 · Dilate
Multiply the outputs by 2: $2 (x - 3)^{2}$ . The vertex is still at $(3, 0)$ , because $2 \cdot 0 = 0$ .
Why this step?
Why dilate before translating down, and not after?
If you subtract 4 first and then multiply by 2, the 4 is doubled too: $2 ((x - 3)^{2} - 4) = 2 (x - 3)^{2} - 8$ , which puts the vertex at $(3, - 8)$ . Translating last keeps the vertex where you put it.
Step 3 · Move it down
Subtract 4: $g (x) = 2 (x - 3)^{2} - 4$ . This is the solid graph in the diagram.
Your step · Calculate

No calculator. Finish it. What is $g (5)$ ?

Your answer

Part 4 Try it

Move the sliders

Lab

Function Grapher

The lab writes every function as $g (x) = a f (b (x + c)) + d$ : $a$ and $d$ act on the outputs, $b$ and $c$ on the inputs. Predict before each slider move, then check.

Open the full Function Grapher lab

Try it · Calculate

No calculator. In the lab, choose the rational family, $f (x) = \frac{1}{x}$ , whose vertical asymptote is $x = 0$ . With $a = 1$ , $b = 1$ and $d = 0$ , what value of $c$ puts the vertical asymptote of $g (x) = f (x + c)$ at $x = - 2$ ? Check your answer with the slider.

Your answer

Part 5 Explain it

Write it the way the exam asks

Nothing you type leaves this page.

Free response is 37.5% of the exam score. This one is shaped like Question 1, Function Concepts.

Free-response practice · Question 1 style

Graphing calculator required. A function $f$ has domain $- 4 \leq x \leq 6$ and range $1 \leq y \leq 5$ . Let $g (x) = f (2 x) - 3$ . Describe the two transformations that take the graph of $f$ to the graph of $g$ , and give the domain and range of $g$ .

Your answer stays on this page. It is never sent or saved.

Model answer

$f (2 x)$ is a horizontal dilation by a factor of $\frac{1}{2}$ , so every input is halved: the domain becomes $- 2 \leq x \leq 3$ , because $2 x$ must be between $- 4$ and 6. Subtracting 3 is a vertical translation down 3 units, so the range becomes $- 2 \leq y \leq 2$ . The horizontal dilation does not change the range, and the vertical translation does not change the domain.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Names the horizontal dilation by a factor of $\frac{1}{2}$ .
Names the vertical translation down 3 units.
Gives the domain of $g$ as $- 2 \leq x \leq 3$ .
Gives the range of $g$ as $- 2 \leq y \leq 2$ .

Part 6 Check yourself

Transforming a table

The two case questions use the same data. Answer them from the table.

Selected values of a function $f$ are below. Let $g (x) = 2 f (x + 1) - 3$ .

Values of $f$
$x$	$f (x)$
−2	3
0	−1
2	5
4	0

Case question 1 · Calculate

No calculator. What is $g (1)$ ?

Your answer

Case question 2 · Multiple choice

No calculator. The point $(4, 0)$ is on the graph of $f$ . Which point is its image on the graph of $g$ ?

The rest stand on their own and do not use the case.

Check yourself · Sort it

No calculator. Tap a transformation of $f$ , then whether it changes the graph vertically or horizontally.

Vertically

Horizontally

Check yourself · Decide and justify

No calculator. The graph of $f$ passes through $(2, 5)$ . Which function's graph passes through $(- 2, 5)$ ?

Which reason supports your answer?

Part 7 Keep it

Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 1.11 Equivalent Representations of Polynomial and Rational Expressions · Fill the statement

No calculator. Tap a blank, then tap a word.

When $f$ is divided by $g$ , $f (x) = g (x) q (x) + r (x)$ , where $q$ is the , $r$ is the , and the degree of $r$ is the degree of $g$ .

From 1.8 Rational Functions and Zeros · Put it in order

No calculator. Tap the steps for solving $r (x) \geq 0$ in the order you do them.

In three lines

Adding outside $f$ translates vertically by $k$ ; adding inside translates horizontally by $- h$ .
Multiplying outside by $a$ dilates vertically by $| a |$ ; multiplying inside by $b$ dilates horizontally by $| \frac{1}{b} |$ ; a negative sign reflects.
Transformations combine from preimage to image, and they can change the domain and the range.

Vocabulary

Additive transformation: Adding a constant to the output, $f (x) + k$ , or to the input, $f (x + h)$ .
Multiplicative transformation: Multiplying the output, $a f (x)$ , or the input, $f (b x)$ , by a nonzero constant.
Translation: A slide of the graph: vertical by $k$ for $f (x) + k$ , horizontal by $- h$ for $f (x + h)$ .
Dilation: A stretch or squeeze: vertical by $| a |$ for $a f (x)$ , horizontal by $| \frac{1}{b} |$ for $f (b x)$ .
Reflection: A flip over the $x$ -axis when $a < 0$ , or over the $y$ -axis when $b < 0$ .
Preimage and image: The set of points before a transformation and the set after it.

Course alignment, for teachers

AP Precalculus topic 1.12, Unit 1: Polynomial and Rational Functions.

Skill 1.C: Construct new functions, using transformations, compositions, inverses, or regressions, that may be useful in modeling contexts, criteria, or data, with and without technology.
Skill 3.A: Describe the characteristics of a function with varying levels of precision, depending on the function representation and available mathematical tools.

This lesson maps all 6 essential knowledge statements of the topic to the part that teaches each one: 6 are checked by a scored question. The map is in the lesson file.

Which way does it move?

Three short ideas, each checked as you go

2a. Translations: adding

2b. Dilations and reflections: multiplying

2c. Combining, and what happens to domain and range

Building a transformed function

Move the sliders

Function Grapher

Write it the way the exam asks

Transforming a table

Bring back earlier ideas

Vocabulary

Labs that use this topic