Transform nonlinear functions

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Lesson progressPractice problems 0/3

Difficulty: Advanced
Estimated time: 30 minutes
Domains: Advanced Math
Techniques: Function-transformationsTranslationsReflectionsStretchesPoint-mapping

What you’ll learn

Spot when a question takes a graph or rule you know and moves or reshapes it.
Write shifts, reflections, and vertical stretches or compressions in function notation.
Find where a named point on $y=f(x)$ lands on the new graph.
Handle the harder inside expression $f(ax+b)$ by solving for the new input.

Prerequisites

Evaluate nonlinear functions and recover inputs0%
Treat the entire expression inside function notation as the input.

Why this matters on the SAT

Tell inside and outside changes apart

The SAT might describe a graph moving in words, give you a new rule like $g(x)=-f(x-3)+1$ , or ask where one named point ends up. All of these come down to one idea:

a change inside $f(\ )$ changes the input, the $x$ -coordinate
a change outside $f(\ )$ changes the output, the $y$ -coordinate

Solution to the example

Take the moves one at a time. Shifting $3$ units right replaces $x$ with $x-3$ . Reflecting across the $x$ -axis multiplies every output by $-1$ . Shifting down $2$ subtracts $2$ outside the function. Put together, that’s

y=-(x-3)^2-2,

so the answer is A.

Choice B falls for a common sign mistake. It’s easy to read $x+3$ as “right $3$ ,” but it moves the graph left. The next section shows why.

How to spot these questions. A graph or rule you already know gets shifted, reflected, stretched, or compressed into a new one. Often the question defines $g$ in terms of $f$ . If no new graph comes out of the old one, nothing is being moved. If the question only asks for a function value, that’s ordinary evaluation. Finding a feature of one quadratic that isn’t being changed, like its vertex, or building a model from a situation, doesn’t move a graph either.

SAT example

The graph of $y=x^2$ is shifted $3$ units to the right, reflected across the $x$ -axis, and shifted $2$ units down. Which equation represents the resulting graph?

A
$y=-(x-3)^2-2$
B
$y=-(x+3)^2-2$
C
$y=(x-3)^2+2$
D
$y=-x^2-5$

Track what happens to one point

A graph is made of points. Transforming it sends every one of those points to a new place, so the question is always the same: where does each point go?

Start with the labeled point in the figure. On the parent graph, the original $y=f(x)$ , the point $P(3,2)$ means $f(3)=2$ . The new graph, called the image, is

g(x)=-2f(x-3)+1.

Transform the coordinates:

3+3=6

and

-2(2)+1=-3

Take the two coordinates one at a time.

The $x$ -coordinate. Whatever goes into $f$ still has to be $3$ , because $f(3)=2$ is the fact you’re carrying over. So the new $x$ has to make

x-3=3,

which gives $x=6$ . On the graph, the point moves $3$ units right.

The $y$ -coordinate. Start with the old output, $2$ , and do what’s outside $f$ :

-2(2)+1=-3.

The negative sign reflects the output across the $x$ -axis, to the other side. The $2$ doubles its distance from that axis, and only then does the $+1$ shift the result up $1$ . So $P(3,2)$ lands at

\boxed{(6,-3)},

the point $P'$ in the figure. That’s four changes in one rule: right $3$ , a reflection, a stretch by $2$ , and up $1$ .

The same two steps work for any point. Call the old point $(u,v)$ : $u$ is the old input, $v$ is the old output, and $f(u)=v$ . For

g(x)=A f(x-h)+k,

the input inside $f$ still has to be $u$ , so $x-h=u$ , which gives $x=u+h$ . Outside, $v$ becomes $Av+k$ . So the point $(u,v)$ lands at

(u+h,\ Av+k).

In short: solve the inside, then apply the outside.

How common changes move a point $(u,v)$

New rule	Where the point goes	What you see
$f(x-h)$	$(u+h,v)$	right $h$ when $h>0$
$f(x)+k$	$(u,v+k)$	up $k$ when $k>0$
$-f(x)$	$(u,-v)$	reflection across the $x$ -axis
$A f(x)$	$(u,Av)$	vertical stretch if $

Try it yourself:

For $g(x)=-2f(x-3)+1$ , make two calls before you calculate. Does a point move left or right? And does a point with a positive original output end up above or below the line $y=1$ ? Then work out the point to check yourself.

When Desmos helps: the whole new graph

For one named point, mapping by hand is quick and exact, and it shows you why the point moves. But when you know the parent’s rule and the question asks about the whole new graph, like its vertex, Desmos may be faster. Define the parent first, then define the image from it:

f(x)=(x-1)^2/2
g(x)=-2f(x-3)+1

Desmos draws the whole image, and you never have to multiply out the new rule. Select the vertex of $g$ and you’ll see $(4,1)$ . The point $(6,-3)$ is on the graph too, right where the hand mapping put it. So for a vertex, a minimum or maximum, an intercept, or a function value as a number, these two lines can get you the answer directly.

Try it yourself:

Change x-3 to x+2. Which way should the graph move? Predict first, then look, then reset the example.

Calculator loads as you approach

Select the vertex of the new graph. Then edit the inside expression and watch the whole image move.

Write the transformed equation

Now flip the job around: the question describes the moves in words, and you write the rule. Horizontal changes go inside $f$ , with the input. Vertical changes go outside, on the output:

g(x)=A f(x-h)+k.

Read the pieces in this order:

Left or right: $x-h$ shifts right $h$ , and $x+h$ shifts left $h$ .
Reflect and stretch: a negative $A$ reflects the graph across the $x$ -axis. The size of $A$ , written $|A|$ , sets the vertical stretch or compression.
Up or down: $+k$ shifts up $k$ , and $-k$ shifts down $k$ .

If you know the parent’s equation, put the whole new input into it first, then apply the outside changes.

Here’s one worked through. Let

f(x)=x^3+2.

Shift its graph $4$ units left, compress it vertically by a factor of $\frac12$ , reflect it across the $x$ -axis, and shift it down $1$ . In function notation, that’s

g(x)=-\frac12 f(x+4)-1.

To write it without $f$ , put $x+4$ into the whole parent rule, $+2$ included:

\begin{aligned} g(x) &=-\frac12\left((x+4)^3+2\right)-1\\[1.4em] &=-\frac12(x+4)^3-2. \end{aligned}

The $-\frac12$ multiplies the $+2$ too, which gives an extra $-1$ . Both forms describe the same graph, but the one written with $f$ makes the moves easier to read.

Check your understanding:

The function $p$ is defined by $p(x)=\sqrt{x+1}$ . Write a function $q$ whose graph is the graph of $p$ shifted $5$ units right, stretched vertically by a factor of $3$ , and shifted $2$ units up.

Common mistake:

Outside, the signs do what they say: $+2$ means up $2$ . Inside, they look backward, because the new input has to solve $x-h=u$ . Say you wrote $f(x+3)$ for a shift right $3$ . Test the parent input $u=0$ . After a shift right, the new graph should show the parent’s output $f(0)$ at $x=3$ . But $f(x+3)$ puts it at $x=-3$ , since $-3+3=0$ .

Example: where does a named point land?

Worked example

The point $A(-1,4)$ lies on the graph of $y=f(x)$ . Which choice gives the corresponding point on the graph of

y=-3f(x-2)+5?

A
$(-3,-7)$
B
$(1,17)$
C
$(1,-7)$
D
$(3,-17)$

Step 1

Say what the point tells you

Since $A(-1,4)$ is on $y=f(x)$ ,

f(-1)=4.

So the old input is $u=-1$ , and the old output is $v=4$ .

Step 2

Find the new input

Whatever goes into $f$ has to equal the old input, $-1$ . So the new $x$ has to solve

x-2=-1,

which gives

x=1.

The point moves $2$ units right, from $-1$ to $1$ .

Step 3

Change the output

Start from the old output, $4$ , and do what’s outside $f$ :

-3(4)+5=-12+5=-7.

The $-3$ reflects the old output and stretches it to three times its distance from the $x$ -axis. Then the $+5$ shifts the graph up $5$ .

Step 4

Name the point and check it

The new point is

\boxed{(1,-7)}.

To check, put $x=1$ into the new rule:

-3f(1-2)+5=-3f(-1)+5=-3(4)+5=-7.

The check runs the mapping backward: $1-2$ lands right back on the old input, $-1$ . The answer is C.

Harder inside changes

So far the inside has been simple, like $x-3$ . A harder SAT question might put more inside $f$ , like this:

g(x)=4-2f(7-3x).

This looks like a big jump, but you already have the tool for it: whatever goes into $f$ still has to equal the old input. The only new part is a slightly longer equation.

Say the point $(5,-2)$ is on $y=f(x)$ , so $f(5)=-2$ . The pieces are out of their usual order, so sort them first. The $7-3x$ is inside $f$ . The $-2$ multiplies the output. The $4$ out front is added last, so it shifts the graph up $4$ .

The $x$ -coordinate. Set the whole inside equal to the old input, $5$ , and solve:

7-3x=5 \quad\Longrightarrow\quad x=\frac23.

The $y$ -coordinate. Start from the old output, $-2$ , and do what’s outside $f$ :

4-2(-2)=8.

So $(5,-2)$ lands at

\boxed{\left(\frac23,8\right)}.

Notice that you never had to work out how far the graph shifted or which way it flipped. Solving the inside took care of all of it.

If a question asks what the graph itself does, the inside $7-3x$ tells you that too:

The $3$ squeezes horizontal distances to $\frac13$ of their size.
The minus sign swaps left and right. That’s a reflection across the $y$ -axis, and it happens before the graph shifts.
The parent’s input $0$ lands where the inside is $0$ . Solving $7-3x=0$ gives $x=\frac73$ .

The same steps work for any inside of the form $ax+b$ , as long as $a\ne0$ . Written in the usual order, the rule is $g(x)=A f(ax+b)+k$ . Our example is $g(x)=-2f(-3x+7)+4$ , so $A=-2$ , $a=-3$ , $b=7$ , and $k=4$ .

Take any point $(u,v)$ on $y=f(x)$ . You solve $ax+b=u$ , just as you solved $7-3x=5$ , and get $x=\frac{u-b}{a}$ . The output still becomes $Av+k$ . So the point lands at

\left(\frac{u-b}{a},\ Av+k\right).

The graph facts carry over the same way. Horizontal distances get multiplied by $\frac1{|a|}$ , like the $\frac13$ in the example. A negative $a$ swaps left and right. And the parent’s input $0$ lands where $ax+b=0$ , which is at $x=-\frac ba$ .

Check your understanding:

If $(8,3)$ lies on $y=f(x)$ and $g(x)=1+4f(2x-6)$ , what point on $y=g(x)$ corresponds to it?

Common mistake:

With $7-3x$ inside, it’s tempting to call it a shift of $7$ , or to divide the old coordinate by $3$ right away. Instead, write one equation, $7-3x=u$ , with $u$ as the old input, and solve it. It handles the reversal, the scaling, and the shift in the right order.

By hand or with Desmos?

Look at what the question asks for. That tells you which way to go.

Use the point map by hand when…

the question gives one point, like a vertex, an intercept, or $f(-1)=4$ , and asks exactly where it lands.
you don’t know the rule for $f$ , as when you’re told only that $A(-1,4)$ is on its graph, so there’s nothing to enter.
the question describes the moves in words, like “shift left $4$ and up $3$ ,” and you have to write the exact equation.
the inside is something like $7-3x$ , and one short equation gives you the new input.

Let Desmos lead when…

you know the rule for $f$ , and the question asks for a number from the new graph, like its vertex, a minimum or maximum, an intercept, or a function value.
the parent’s rule is written out long, like $x^2-8x+20$ , so the point you need is hard to see until you graph the new function.
the answer choices are whole new rules, and you can compare them quickly by seeing which graphs overlap.

Either way, answer exactly what the question asks for. In Desmos, select or evaluate it on the new graph, not the parent. If you get a decimal, like $3.5$ , and the choices are exact, convert it to $\frac72$ .

You won’t need regression or Log Mode for these. A transformation changes a rule or graph you already know. It doesn’t build a new model from data.

Practice problems

Three problems to try on your own: write an equation, map a point, then find a vertex through a harder inside change. The last one is the toughest, so take your time with it. Before each one, look at what it asks for and decide: by hand or Desmos?

Write a combined transformation

Practice problem

The function $f$ is defined by

f(x)=x^3-2.

The graph of $y=g(x)$ is obtained by shifting the graph of $y=f(x)$ $4$ units left, reflecting it across the $x$ -axis, vertically compressing it by a factor of $\frac12$ , and shifting it $3$ units up. Which equation defines $g$ ?

Calculator loads as you approach

Use Desmos if comparing the parent and new graphs helps.

Map a point through both coordinates

Practice problem

The point $P(-2,5)$ lies on the graph of $y=f(x)$ . Which choice gives the corresponding point on the graph of

y=3f(x+4)-2?

Calculator loads as you approach

There’s no rule for

f

to graph here, so map the point by hand.

Find a transformed quadratic vertex

Practice problem

Let

f(x)=x^2-8x+20.

A second function is defined by

g(x)=5-3f(11-2x).

Which choice gives the coordinates of the vertex of the graph of $y=g(x)$ ?

Calculator loads as you approach

Define f and g, select the vertex of g, and match its decimals to an exact choice.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Solve the inside, then apply the outside. In $g(x)=A f(x-h)+k$ , that sends a point $(u,v)$ to $(u+h,Av+k)$ .
Inside signs look backward because the new input has to make $x-h=u$ .
A negative outside factor reflects the graph across the $x$ -axis. Its size, $|A|$ , gives the vertical stretch or compression.
For $g(x)=A f(ax+b)+k$ , set the whole inside equal to the old input, $ax+b=u$ , and solve instead of guessing the horizontal move.
For one named point, mapping by hand is usually the quickest exact way. When you know the parent’s rule and need a number from the whole new graph, defining $f$ and $g$ in Desmos can get you the answer directly.

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