Math Core

You already know that $f(x - h) + k$ slides a graph and that a negative sign flips it. Precalculus asks more of you: transforming functions you only know from a graph or a few points, handling a horizontal stretch and a shift in the same expression, and predicting exactly how the domain and range move. The key is to treat every transformation as a rule that moves points, and to keep careful track of the order.

One rule for every transformation

The general transformation

Write the new function in the form

g(x) = a\,f\big(b(x - h)\big) + k

Then every point $(x, y)$ on the graph of $f$ moves to

\left(\frac{x}{b} + h,\ a y + k\right)

on the graph of $g$ . The inside numbers $b$ and $h$ act on $x$ -coordinates; the outside numbers $a$ and $k$ act on $y$ -coordinates.

Why divide by $b$ ? On $g$ , the input $x$ gets multiplied by $b$ before $f$ sees it. To reach an input that $f$ already knew, say $x_0$ , you need $bx = x_0$ , so $x = \dfrac{x_0}{b}$ . The inside of a function always works backward, which is also why $f(x - 3)$ moves right instead of left.

Here is what each constant does on its own:

Constant	Effect on the graph of $f$
$a$	vertical stretch by $\lvert a \rvert$ (compression if $\lvert a \rvert < 1$ ); also reflect across the $x$ -axis if $a < 0$
$b$	horizontal compression by $\dfrac{1}{\lvert b \rvert}$ (stretch if $\lvert b \rvert < 1$ ); also reflect across the $y$ -axis if $b < 0$
$h$	shift right $h$ (left if $h < 0$ )
$k$	shift up $k$ (down if $k < 0$ )

Order matters

Vertical transformations follow the usual order of operations on the output: first multiply by $a$ , then add $k$ . If the steps happen in a different order, the equation changes.

Worked example: Stretch then shift, or shift then stretch?

Start with $f(x) = \sqrt{x}$ .

Stretch vertically by $3$ , then shift up $2$ .
Shift up $2$ , then stretch vertically by $3$ .

1. Stretching gives $3\sqrt{x}$ . Shifting adds $2$ to that: $g(x) = 3\sqrt{x} + 2$ . The starting point $(0, 0)$ moves to $(0, 2)$ .

2. Shifting gives $\sqrt{x} + 2$ . Stretching multiplies the whole output by $3$ : $g(x) = 3(\sqrt{x} + 2) = 3\sqrt{x} + 6$ . The starting point moves to $(0, 6)$ .

The parent (dashed), 3√x + 2, and 3√x + 6. Same steps, different order, different graphs.Open in grapher →

The same care applies inside the function. The horizontal steps in $f\big(b(x - h)\big)$ happen in this order: stretch or compress by $b$ first, then shift by $h$ . The shift has to be read off after factoring out $b$ . (You could also shift first and compress second, but then the shift amount changes: for $f(2x - 6)$ that would be right $6$ , then compress. Precalculus uses the factored, compress-first reading, because it lets you use the point rule above.)

Common mistake

When you compress first, the horizontal shift in $g(x) = f(2x - 6)$ is not $6$ . Factor the inside first: $2x - 6 = 2(x - 3)$ , so $g(x) = f\big(2(x - 3)\big)$ . The graph is compressed horizontally by $\dfrac{1}{2}$ and then shifted right 3. Reading "right 6" off $2x - 6$ is the most common transformation mistake in precalculus.

Worked example: A horizontal compression with a shift

The point $(4, 5)$ lies on the graph of $f$ . Find the corresponding point on $g(x) = f(2x - 6)$ , and check it.

Factor: $g(x) = f\big(2(x - 3)\big)$ , so $b = 2$ , $h = 3$ , $a = 1$ , $k = 0$ .

Apply $(x, y) \to \left(\dfrac{x}{b} + h,\ ay + k\right)$ :

(4, 5) \to \left(\frac{4}{2} + 3,\ 5\right) = (5, 5)

Check: $g(5) = f(2 \cdot 5 - 6) = f(4) = 5$ . The point $(5, 5)$ is on $g$ .

Tip

You can always find a transformed point by solving instead of memorizing. For $g(x) = f(2x - 6)$ , ask: which $x$ makes the inside equal to $4$ ? Solve $2x - 6 = 4$ to get $x = 5$ . Then $g(5) = f(4) = 5$ .

How domain and range transform

Because the inside numbers only move $x$ -coordinates and the outside numbers only move $y$ -coordinates, you can transform a domain and range without a graph at all.

Push each domain endpoint through $x \to \dfrac{x}{b} + h$ .
Push each range endpoint through $y \to a y + k$ .
If $a$ or $b$ is negative, the endpoints swap order, so rewrite the interval from smallest to largest.

Worked example: Transforming a domain and range

A function $f$ has domain $[-2, 6]$ and range $[-1, 3]$ . Find the domain and range of $g(x) = -2 f\left(\dfrac{x}{2} + 1\right) + 4$ .

Factor the inside: $\dfrac{x}{2} + 1 = \dfrac{1}{2}(x + 2)$ . So $b = \dfrac{1}{2}$ , $h = -2$ , $a = -2$ , $k = 4$ .

Domain. The rule is $x \to \dfrac{x}{1/2} - 2 = 2x - 2$ :

-2 \to -6, \qquad 6 \to 10

The domain of $g$ is $[-6, 10]$ . (Check by solving: $-2 \le \dfrac{x}{2} + 1 \le 6$ gives $-6 \le x \le 10$ .)

Range. The rule is $y \to -2y + 4$ :

-1 \to 6, \qquad 3 \to -2

The endpoints swapped because $a < 0$ , so the range of $g$ is $[-2, 6]$ .

Graphing with key points

For a familiar parent function, pick a few key points, push them through the rule, and connect them with the parent's shape. A reflection across the $y$ -axis is part of $b$ , so factor out the negative too.

Worked example: A reflected square root

Graph $g(x) = 2\sqrt{3 - x} - 1$ and state its domain and range.

Rewrite the inside: $3 - x = -(x - 3)$ , so $g(x) = 2\sqrt{-(x - 3)} - 1$ with $b = -1$ , $h = 3$ , $a = 2$ , $k = -1$ .

The graph of $\sqrt{x}$ is reflected across the $y$ -axis, shifted right $3$ , stretched vertically by $2$ and shifted down $1$ . Points move by $(x, y) \to (-x + 3,\ 2y - 1)$ :

On $\sqrt{x}$	On $g$
$(0, 0)$	$(3, -1)$
$(1, 1)$	$(2, 1)$
$(4, 2)$	$(-1, 3)$

y = sqrt(x)y = 2sqrt(3 - x) - 1(3, -1)(2, 1)(-1, 3)Open in grapher →

The graph starts at $(3, -1)$ and heads left and up. Domain: $3 - x \ge 0$ , so $x \le 3$ . Range: $y \ge -1$ .

Check one point: $g(-1) = 2\sqrt{4} - 1 = 3$ . It matches.

Notice that a reflection across the $y$ -axis has no effect on an even function, since $f(-x) = f(x)$ , and for an odd function it gives the same graph as a reflection across the $x$ -axis, since $f(-x) = -f(x)$ . For example, $(-x)^3 = -x^3$ .

Practice

Practice 1

The point $(6, -2)$ is on the graph of $f$ . What point must be on the graph of $g(x) = f(x + 4) - 3$ ?

Enter a point like (2, -3)

Practice 2

The point $(6, -2)$ is on the graph of $f$ . What point must be on the graph of $g(x) = f(3x)$ ?

Enter a point like (2, -3)

Practice 3

The point $(6, -2)$ is on the graph of $f$ . What point must be on the graph of $g(x) = -f(2x - 4) + 1$ ?

Enter a point like (2, -3)

Practice 4

Which describes the graph of $g(x) = f(2x + 6)$ compared with the graph of $f$ ?

Practice 5

Start with $f(x) = x^2$ . Stretch the graph horizontally by a factor of $2$ , then shift it right $1$ and up $3$ . Write the new function $g(x)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 6

A function $f$ has domain $[0, 8]$ . What is the domain of $g(x) = f(4 - x)$ ? Write it as a compound inequality.

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 7

A function $f$ has range $[-3, 5]$ . What is the range of $g(x) = -3f(x) + 2$ ? Write it as a compound inequality in $y$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 8

The graph shown is a transformation of $f(x) = \sqrt{x}$ . It starts at $(2, 1)$ and passes through $(-2, 3)$ . Write its equation.

y = sqrt(2 - x) + 1(2, 1)(-2, 3)Open in grapher →

Enter an expression, e.g. 3x^2 - 2x + 1