ACT Math · Functions · free lesson

Transformations of Graphs

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

The topic in one minute

Press Play: the idea and a worked example, explained aloud (about 1:16).

Transformations of Graphs

Start with the graph of $y = f(x)$ .

Outside the function, like $f(x) + 3$ or $2f(x)$ : the outputs change, so the graph moves up and down.
Inside the parentheses, like $f(x - 2)$ or $f(2x)$ : the inputs change, so the graph moves left and right.

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Read what is said

A small change to a rule can move a whole graph, flip it or stretch it. The test asks you to connect the change in the rule with the change in the graph, so first ask where the change was made.
Watch the V drawn in. The parent graph has its corner at the origin. The image has its corner at two, three, so it moved right two, and up three.
Here is the rule. The number outside does what its sign says. The number inside does the opposite of what its sign suggests, so minus two inside means right.
Now an example. The graph of y equals x plus four, squared, minus three, is the graph of x squared moved which way? Follow the lowest point. It comes where the inside is zero.
So the lowest point moves from the origin to negative four, negative three. Look at the red curve: it sits four units to the left of the parent, and three units lower.
Here is where points are lost. The wrong choices are the ones that read the sign inside as it looks, or take the outside number as a move to the side. Also, never treat a stretch as a shift.
Remember: decide first whether the change is outside or inside. Outside acts on the outputs, up and down. Inside acts on the inputs, and does the opposite of its sign. Now practice it with new numbers.

The idea

A new graph 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. The ACT 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 $f(x) + 3$ or $2f(x)$ , is made to the outputs, so it acts up and down. A change inside the parentheses, such as $f(x - 2)$ or $f(2x)$ , is made to the inputs, so it acts left and right.

Shifts

Take $f(x) = \lvert x \rvert$ , whose graph is a V with its corner at $(0, 0)$ . Let $g(x) = f(x) + 3$ and $h(x) = f(x - 2)$ . Compare the outputs for the same inputs:

\begin{aligned} f(x) &= \left|x\right| \\ g(x) &= f(x) + 3 \\ h(x) &= f(x - 2) \end{aligned}

\begin{aligned} y &= f(x) \\ &= \left|x\right| \\ &= [1, 0, 1, 2, 3] \\ y &= g(x) \\ &= \left|x\right| + 3 \\ &= [4, 3, 4, 5, 6] \\ y &= h(x) \\ &= \left|x - 2\right| \\ &= [3, 2, 1, 0, 1] \end{aligned}

Every output of $g$ is 3 more than the output of $f$ : the graph has gone up 3. The outputs of $h$ are those of $f$ arriving 2 places later: the corner, where the output is 0, comes at $x = 0$ for $f$ and at $x = 2$ for $h$ . So the graph of $h$ has gone 2 units to the right, although its rule shows a minus sign.

Rule.

For $k > 0$ : $y = f(x) + k$ is the graph of $f$ moved up $k$ , and $y = f(x) - k$ moved down $k$ . For $h > 0$ : $y = f(x - h)$ is the graph of $f$ moved right $h$ , and $y = f(x + h)$ moved left $h$ . A change inside the parentheses does the opposite of what its sign suggests.

Click the picture to watch the parent graph $y = \lvert x \rvert$ and its image $y = \lvert x - 2 \rvert + 3$ drawn in together. The corner moves from $(0, 0)$ to $(2, 3)$ : right 2 and up 3.

In the animation the inside number $h$ grows from $-4$ to 4. The V of $y = \lvert x - h \rvert$ slides from left to right, and its corner is always at $x = h$ .

Reflections

A minus sign outside, $y = -f(x)$ , changes the sign of every output: the graph flips over the $x$ -axis. A minus sign inside, $y = f(-x)$ , changes the sign of every input: the graph flips over the $y$ -axis. The picture uses $f(x) = \sqrt{x}$ , which leans to one side, so the two flips are easy to tell apart.

Stretches

Multiplying outside, $y = a \cdot f(x)$ , multiplies every output by $a$ : for $a > 1$ the graph is stretched away from the $x$ -axis, and for $0 < a < 1$ it is squeezed toward it. Multiplying inside, $y = f(bx)$ , acts on the inputs, and again the opposite way: for $b > 1$ the graph is squeezed toward the $y$ -axis by the factor $\frac{1}{b}$ . Follow the top point $(0, 4)$ and the zero $(2, 0)$ of $f(x) = 4 - x^{2}$ :

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

For $y = 2f(x)$ the top goes up to $(0, {8})$ , but the zero stays at $(2, 0)$ , because twice 0 is still 0. For $y = f(2x)$ the zero comes where $2x = 2$ , at $x = 1$ : the graph is half as wide and just as tall.

All of them at once

Drag the sliders to change $y = a \cdot f(x - h) + k$ for $f(x) = \lvert x \rvert$ . The dashed V is the parent. The corner of the image is always at $(h, k)$ ; the factor $a$ sets how steep the arms are, and a negative $a$ turns the V upside down.

To follow one point, act on each coordinate. A point $(p, q)$ of $y = f(x)$ goes to $(p + h, aq + k)$ on $y = a \cdot f(x - h) + k$ . For the point $(2, 3)$ and $y = 2f(x + 1) - 4$ , here $h$ is $-1$ :

\begin{aligned} X &= 2 + (-1) \\ &= 1 \\ Y &= 2 \times 3 - 4 \\ &= 2 \end{aligned}

The point goes to $({1}, {2})$ : left 1, then the height doubled and lowered by 4.

Warning.

The slips that cost the most points: moving the graph left for $f(x - 1)$ , when it moves right; reading $f(x) - k$ as a move to the left; and mixing up a stretch with a shift, as if $2f(x)$ moved the graph up 2.

Note.

On the ACT, these questions ask which way a graph has moved, the rule after a given move, where a point goes, which graph matches an equation, and what a flip or a stretch does. Moving whole figures in the plane, by rotations and reflections, is a topic of its own (topic 40).

Worked examples

In each example, first decide whether the change is outside the function (up or down, a flip over the $x$ -axis, a stretch from the $x$ -axis) or inside the parentheses (left or right, a flip over the $y$ -axis, a squeeze toward the $y$ -axis). The numbers in the working are found by the engine.

Example.

1. Naming the moves. The graph of $y = (x + 4)^{2} - 3$ is the graph of $y = x^{2}$ moved in which way?

Write the rule as $f(x + 4) - 3$ with $f(x) = x^{2}$ . Follow the lowest point $(0, 0)$ of $y = x^{2}$ . On the new graph the lowest point comes where the inside is 0, and its height is 0 less 3:

\begin{aligned} x + 4 &= 0 \\ x &= -4 \\ y &= 0 - 3 \\ &= -3 \end{aligned}

The lowest point has gone from $(0, 0)$ to $({-4}, {-3})$ : the graph moved 4 units to the left and 3 units down. The ACT's wrong choices are "4 to the right" (reading the sign inside as it looks) and "3 to the left" (taking the outside number as a move to the side).

Example.

2. Writing the rule after a move. The function $f$ is defined by $f(x) = x^2 - 2x + 5$ . The graph of $y = g(x)$ is the graph of $y = f(x)$ moved 3 units to the right and 1 unit up. Write $g(x)$ in the form $x^{2} + bx + c$ .

Right 3 is a change inside: replace every $x$ by $x - 3$ . Up 1 is a change outside: add 1 to the whole rule. So $g(x) = f(x - 3) + 1$ :

\begin{aligned} y &= (x - 3)^{2} - 2(x - 3) + 5 + 1 \\ &= x^{2} - 8x + 21 \end{aligned}

Check with the lowest point. That of $f$ is at $x = 1$ ; that of $g$ should be at $x = 1 + 3$ , one unit higher:

\begin{aligned} f(x) &= x^{2} - 2x + 5 \\ g(x) &= f(x - 3) + 1 \\ p &= f(1) \\ &= 1^{2} - 2 \times 1 + 5 \\ &= 4 \\ q &= g(1 + 3) \\ &= (1 + 3 - 3)^{2} - 2 \times (1 + 3 - 3) + 5 + 1 \\ &= 5 \end{aligned}

The lowest point goes from $(1, {4})$ to $({4}, {5})$ : right 3 and up 1. A common slip replaces only the $x$ in $x^{2}$ and forgets the $-2x$ .

Example.

3. Two flips. The point $(5, 4)$ is on the graph of $f(x) = \sqrt{x - 1} + 2$ . Write the rules of $y = -f(x)$ and $y = f(-x)$ , and say where the point goes on each.

$-f(x)$ changes the sign of the whole output, so $-f(x) = -\sqrt{x - 1} - 2$ ; the point goes to $(5, -4)$ . $f(-x)$ puts $-x$ in place of $x$ , so $f(-x) = \sqrt{-x - 1} + 2$ ; the point goes to $(-5, 4)$ . Check each by putting the new point into its rule:

\begin{aligned} L &= -(\sqrt{5 - 1} + 2) \\ &= -4 \\ R &= \sqrt{-(-5) - 1} + 2 \\ &= 4 \end{aligned}

$-f(5)$ is $-4$ and $f(-(-5))$ is $4$ , as they should be. Flipping over the $x$ -axis changes $y$ ; flipping over the $y$ -axis changes $x$ .

Example.

4. A stretch outside and a squeeze inside. The point $(6, 4)$ is on the graph of $y = f(x)$ . Which point must be on $y = 3f(x)$ ? Which point must be on $y = f(3x)$ ?

Outside, the 3 multiplies the output: the $x$ -coordinate stays and the $y$ -coordinate is tripled. Inside, the 3 multiplies the input, so the new graph reaches the same height where $3x$ is 6:

\begin{aligned} y &= 3 \times 4 \\ &= 12 \\ 3x &= 6 \\ x &= 2 \end{aligned}

$(6, {12})$ is on $y = 3f(x)$ , a vertical stretch by 3. $({2}, 4)$ is on $y = f(3x)$ , a horizontal squeeze by the factor $\frac{1}{3}$ . The usual trap is $({18}, 4)$ , which treats the inside 3 as a stretch away from the $y$ -axis.

Example.

5. Matching a graph with its rule. Which equation has the graph shown: $y = -\lvert x + 2 \rvert + 3$ , $y = -\lvert x - 2 \rvert + 3$ , $y = \lvert x + 2 \rvert + 3$ or $y = -\lvert x + 3 \rvert + 2$ ?

The V opens downward, so the parent $\lvert x \rvert$ has been flipped: there is a minus sign outside. The corner is at $(-2, 3)$ : moved left 2 (inside, $x + 2$ ) and up 3 (outside, $+3$ ). That is $y = -\lvert x + 2 \rvert + 3$ . Check the corner and one more point of the graph, $(0, 1)$ :

\begin{aligned} c &= -\left|-2 + 2\right| + 3 \\ &= 3 \\ y &= -\left|0 + 2\right| + 3 \\ &= 1 \end{aligned}

Both agree with the picture. $-\lvert x - 2 \rvert + 3$ has its corner at $x = 2$ ; $\lvert x + 2 \rvert + 3$ opens upward; $-\lvert x + 3 \rvert + 2$ swaps the two numbers.

Example.

6. A shift in a story. A ball thrown from the ground is at height $h(t) = -16t^2 + 48t$ feet after $t$ seconds. A second ball is thrown 1 second later from a balcony 5 feet up, in the same way. Its height is $g(t) = h(t - 1) + 5$ . When does the second ball reach its highest point, and how high is it?

The first ball is highest halfway between its zeros, at $t = \frac{48}{32}$ . The second ball's graph is the first one moved right 1 (it starts 1 second later) and up 5:

\begin{aligned} h(t) &= -16t^{2} + 48t \\ T &= \frac{48}{32} \\ &= \frac{3}{2} \\ m &= h(T) \\ &= -16 \times (\frac{3}{2})^{2} + 48 \times \frac{3}{2} \\ &= 36 \\ s &= T + 1 \\ &= \frac{3}{2} + 1 \\ &= \frac{5}{2} \\ M &= m + 5 \\ &= 36 + 5 \\ &= 41 \end{aligned}

The second ball is highest at $t = \frac{5}{2}$ seconds, at $41$ feet. Written out, the rule of the second ball is:

\begin{aligned} g &= -16(t - 1)^{2} + 48(t - 1) + 5 \\ &= -16t^{2} + 80t - 59 \end{aligned}

Practice

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

Practise transformations of graphs with new numbers