SAT Math · Advanced Math · free lesson

Function Transformations

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

A new function from an old one

Start with the graph of $y = f(x)$ . A small change to the rule moves the whole graph, flips it or stretches it, without changing its shape in any other way. The test asks you to connect the change in the rule with the change in the graph.

There are two places where a change can be made. A change outside the function, such as $f(x) + 3$ , is made to the outputs, so it moves the graph up or down. A change inside the parentheses, such as $f(x - 2)$ , is made to the inputs, so it moves the graph left or right.

Shifts up and down

Take $f(x) = x^{2}$ . The function $g(x) = f(x) + 3$ adds 3 to every output. Every point of the graph goes up 3 units:

\begin{aligned} f(x) &= x^{2} \\ g(x) &= f(x) + 3 \end{aligned}

\begin{aligned} y &= f(x) \\ &= x^{2} \\ &= [4, 1, 0, 1, 4] \\ y &= g(x) \\ &= x^{2} + 3 \\ &= [7, 4, 3, 4, 7] \end{aligned}

Rule.

The graph of $y = f(x) + k$ is the graph of $f$ moved up $k$ units. The graph of $y = f(x) - k$ is the graph of $f$ moved down $k$ units.

Shifts left and right

Now take $h(x) = f(x - 2)$ , which is $(x - 2)^{2}$ . Compare the outputs of $f$ and $h$ for the same inputs:

\begin{aligned} h(x) &= f(x - 2) \end{aligned}

\begin{aligned} y &= f(x) \\ &= x^{2} \\ &= [4, 1, 0, 1, 4, 9, 16] \\ y &= h(x) \\ &= (x - 2)^{2} \\ &= [16, 9, 4, 1, 0, 1, 4] \end{aligned}

The outputs of $h$ are the outputs of $f$ , arriving 2 places later. The lowest value, 0, comes at $x = 0$ for $f$ and at $x = 2$ for $h$ . So the graph has moved 2 units to the right, although the rule shows a minus sign.

Rule.

The graph of $y = f(x - h)$ is the graph of $f$ moved right $h$ units. The graph of $y = f(x + h)$ is the graph of $f$ moved left $h$ units. A change inside the parentheses does the opposite of what its sign suggests.

Watch the graph of $y = f(x - h)$ slide to the right as $h$ grows from 0 to 4. The gray curve is $f$ itself.

Flips and stretches

Multiplying changes the graph in a different way.

New rule	What happens to the graph of $f$	A point $(p, q)$ goes to
$y = -f(x)$	flipped over the $[-2, -1, 0, 1, 2, 3, 4]$ -axis	$(p, -q)$
$y = f(-x)$	flipped over the $[16, 9, 4, 1, 0, 1, 4]$ -axis	$(-p, q)$
$y = a \cdot f(x)$ , with $a > 1$	stretched away from the $[-2, -1, 0, 1, 2, 3, 4]$ -axis	$(p, aq)$
$y = a \cdot f(x)$ , with $0 < a < 1$	squeezed toward the $[-2, -1, 0, 1, 2, 3, 4]$ -axis	$(p, aq)$

The picture uses $f(x) = \sqrt{x}$ , whose graph leans to one side, so the two flips are easy to tell apart.

All of them at once

Drag the sliders to change $y = a \cdot f(x - h) + k$ for $f(x) = x^{2}$ . The dashed curve is $f$ . The red point is where the lowest point $(0, 0)$ of $f$ has gone: it is always at $(h, k)$ .

Going backward

Sometimes the graph you are shown is already the changed one. If the graph shown is $y = f(x) + 4$ and the question asks for $f$ , the shown graph is 4 units too high. Take the 4 away to get back to $f$ . Do not add it a second time.

Warning.

Three slips cost the most points here. Choosing a left or right shift when the rule shows an up or down shift. Moving the graph left for $f(x - 1)$ , when it moves right. Adding $k$ again when the graph shown is already $f(x) + k$ .

Worked examples

In each example, first decide whether the change is outside the function (up or down, flip, stretch) or inside the parentheses (left or right). The numbers in the working are found by the engine.

Example.

1. A shift written as a formula. The function $f$ is defined by $f(x) = 2x^2 - 3x + 7$ . The graph of $y = g(x)$ is the graph of $y = f(x)$ moved down 5 units. Which rule defines $g$ ?

Down 5 means every output is 5 less, so $g(x) = f(x) - 5$ . The 5 is taken from the whole rule, outside it:

\begin{aligned} y &= 2x^{2} - 3x + 7 - 5 \\ &= 2x^{2} - 3x + 2 \end{aligned}

So $g(x)$ is $2x^{2} - 3x + 2$ . Only the constant has changed. A wrong choice on the test would be $2(x - 5)^2 - 3(x - 5) + 7$ : that puts the 5 inside, which moves the graph 5 units to the right, not down.

Example.

2. Where a point goes. The point $(2, -1)$ is on the graph of $y = f(x)$ . Which point must be on the graph of $y = f(x) + 3$ ? Which point must be on the graph of $y = f(x + 4)$ ?

Adding 3 outside moves the point up 3: only the $y$ -coordinate changes. Adding 4 inside moves the graph left 4: only the $x$ -coordinate changes, and it goes down, not up.

\begin{aligned} y &= y + 3 \\ &= -1 + 3 \\ &= 2 \\ x &= x - 4 \\ &= 2 - 4 \\ &= -2 \end{aligned}

The point (2, $2$ ) is on $y = f(x) + 3$ , and the point ( $-2$ , $-1$ ) is on $y = f(x + 4)$ . Check the second one: when $x$ is $-2$ , the input of $f$ is $-2$ + 4, which is $2$ , and $f$ ( $2$ ) is $-1$ .

Example.

3. A value of the moved function. The function $f$ is defined by $f(x) = (x - 1)(x + 2)(x - 4)$ . The graph of $y = g(x)$ is the graph of $y = f(x)$ moved up 6 units. What is $g(0)$ ?

Up 6 means $g(x) = f(x) + 6$ . You do not need to multiply the rule out. Find $f(0)$ and add 6.

\begin{aligned} f(x) &= (x - 1)(x + 2)(x - 4) \\ u &= f(0) \\ &= (0 - 1) \times (0 + 2) \times (0 - 4) \\ &= 8 \\ y &= u + 6 \\ &= 8 + 6 \\ &= 14 \end{aligned}

So $g(0)$ is $14$ . Two negative factors make the product positive: watch the signs.

Example.

4. The graph shown is already shifted. The graph of $y = f(x) + 4$ is a line that crosses the $y$ -axis at $(0, 7)$ and has slope 2. Which rule defines $f$ ?

The line shown has the rule $2x + 7$ . That is $f(x) + 4$ , not $f(x)$ . To get $f$ , take the 4 away:

\begin{aligned} y &= 2x + 7 - 4 \\ &= 2x + 3 \\ W &= 2x + 7 + 4 \\ &= 2x + 11 \end{aligned}

So $f(x)$ is $2x + 3$ . Its graph is 4 units below the line shown. The trap, $2x + 11$ , adds the 4 a second time.

Example.

5. Intercepts after a shift to the side. The function $f$ is defined by $f(x) = (x + 3)(x - 2)(x - 5)$ , and $g(x) = f(x - 1)$ . What is the sum of the $x$ -intercepts of the graph of $g$ ?

The $x$ -intercepts of $f$ are where a factor is 0. The engine finds them. The graph of $g$ is the graph of $f$ moved 1 unit to the right, so each intercept grows by 1.

\begin{aligned} (x + 3)(x - 2)(x - 5) &= 0 \\ x &= [-3, 2, 5] \\ x &= x + 1 \\ &= [-2, 3, 6] \\ S &= \operatorname{sum}(x) \\ &= 7 \end{aligned}

The sum is $7$ . To check, write $g$ out: replacing $x$ by $x - 1$ gives $g(x) = (x + 2)(x - 3)(x - 6)$ , and the engine solves it directly:

\begin{aligned} (x + 2)(x - 3)(x - 6) &= 0 \\ x &= [-2, 3, 6] \end{aligned}

Moving the intercepts left instead is the usual mistake. The minus sign inside the parentheses means right.

Example.

6. A flip and a stretch. The function $f(x) = x^2 - 4x$ has its lowest point at $(2, -4)$ . Where is the highest point of the graph of $y = -3f(x)$ , and what is the rule of that function?

Multiplying by $-3$ is done to the outputs. Each $y$ -coordinate is multiplied by 3, which stretches the graph, and changes sign, which flips it over the $x$ -axis. The $x$ -coordinates stay as they are.

\begin{aligned} f(x) &= x^{2} - 4x \\ k &= f(h) \\ &= 2^{2} - 4 \times 2 \\ &= -4 \\ m &= -3k \\ &= -3 \times (-4) \\ &= 12 \\ y &= -3(x^{2} - 4x) \\ &= -3x^{2} + 12x \end{aligned}

The lowest point ( $2$ , $-4$ ) of $f$ becomes the highest point ( $2$ , $12$ ) of the new graph, and the rule is $-3x^{2} + 12x$ . Both graphs cross the $x$ -axis at the same places, because 0 times $-3$ is still 0.

Practice

Answer SAT-style questions on function transformations. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise function transformations with new numbers