Transformations of Functions — AP® Precalculus Notes

Once you know the graph of one function well, you know a whole family. Shift it, stretch it, flip it, and you get new functions whose graphs you can draw without plotting a single point. The original is the parent; the new one is its transformation, written in terms of the parent, such as $g(x) = 2f(x - 3) - 1$ .

Outside changes act on outputs

Anything done to $f(x)$ after the function has acted changes the outputs, so it moves the graph up or down or stretches it vertically. The inputs stay where they were.

Vertical translation: $g(x) = f(x) + k$ moves every point $k$ units up (down if $k < 0$ ).
Vertical dilation: $g(x) = a f(x)$ multiplies every output by $a$ . If $a > 1$ the graph is stretched away from the $x$ -axis; if $0 < a < 1$ it is compressed toward it.
Reflection across the $x$ -axis: $g(x) = -f(x)$ changes the sign of every output. A negative $a$ dilates and reflects at once.

Inside changes act on inputs, in reverse

Anything done to $x$ before the function acts changes which input produces each output. That is why the direction feels backward.

Horizontal translation: $g(x) = f(x - h)$ moves the graph $h$ units right. Check with a value: $g(h) = f(0)$ , so the output $f$ had at $0$ now appears at $h$ .
Horizontal dilation: $g(x) = f(bx)$ multiplies every input by $\frac{1}{b}$ . So $f(2x)$ is squeezed toward the $y$ -axis by a factor of $\frac{1}{2}$ , and $f(\frac{x}{3})$ is stretched by a factor of $3$ .
Reflection across the $y$ -axis: $g(x) = f(-x)$ changes the sign of every input.

Formula	Acts on	The point $(a, b)$ on $f$ moves to
$f(x) + 4$	outputs	$(a, b + 4)$
$3f(x)$	outputs	$(a, 3b)$
$-f(x)$	outputs	$(a, -b)$
$f(x - 4)$	inputs	$(a + 4, b)$
$f(3x)$	inputs	$(\frac{a}{3}, b)$
$f(-x)$	inputs	$(-a, b)$

Combining transformations

For $g(x) = a f(x - h) + k$ , work from the input outward: shift the input, apply $f$ , multiply by $a$ , add $k$ . The order matters on the output side. Multiplying by $2$ and then adding $1$ is not the same as adding $1$ and then multiplying by $2$ .

Worked example 1. Let $f(x) = x^2$ and $g(x) = 2f(x - 3) - 1$ . Where does the point $(1, 1)$ on $f$ go, and what is the vertex of $g$ ?

The input moves $3$ right: $1 \to 4$ . The output is doubled, then lowered by $1$ : $1 \to 2 \to 1$ . So $(1, 1)$ goes to $(4, 1)$ .

Check with the formula: $g(4)$ $= 2f(1) - 1$ $= 2(1) - 1$ $= 1$ . Correct.

The vertex $(0, 0)$ goes to $(0 + 3, 2 \cdot 0 - 1)$ $= (3, -1)$ .

The parent graph f(x) = x^2 with vertex (0, 0) and the transformed graph g(x) = 2f(x - 3) - 1, narrower, with vertex (3, -1); the point (1, 1) on f moves to (4, 1) on g. — **Figure 1.** $f(x) = x^2$ and $g(x) = 2f(x - 3) - 1$ : $3$ right, twice as tall, $1$ down. LevlPrep original diagram.

In the figure, $g$ looks narrower. That is the vertical stretch: each output, measured from the vertex level, is twice as large.

Writing the new formula

To write $g$ as an expression in $x$ , replace every $x$ in $f$ with what is inside the parentheses. Missing one is the most common slip.

Worked example 2. Let $f(x) = x^2 - 4x$ . Write $g(x) = f(x - 1) + 3$ in terms of $x$ .

Replace each $x$ with $(x - 1)$ , in parentheses: $f(x - 1)$ $= (x - 1)^2 - 4(x - 1)$ .

Add $3$ : $g(x)$ $= (x - 1)^2 - 4(x - 1) + 3$ .

Expand if you want standard form: $x^2 - 2x$ $+ 1 - 4x$ $+ 4$ $+ 3$ $= x^2 - 6x$ $+ 8$ .

How the domain and range change

Because inside changes act on inputs, they change the domain. Because outside changes act on outputs, they change the range.

Worked example 3. $f$ has domain $[0, 4]$ and range $[-1, 3]$ . Find the domain and range of $g(x) = -2f(x - 1)$ .

Domain: $g(x)$ needs $x - 1$ in $[0, 4]$ , so $1 \le x \le 5$ . The domain is $[1, 5]$ .

Range: multiply the ends by $-2$ : $-1 \to 2$ and $3 \to -6$ . Reorder from least to greatest: $[-6, 2]$ .

Tables work the same way. If $n(x) = k(x - 3)$ and the table gives $k(2) = 6$ , then $n(5) = k(2) = 6$ . To find where $n(x) = 6$ , solve $x - 3 = 2$ .