Combining Transformations

Real graphs often use several transformations at once, like $y = -2\sqrt{x + 3} + 4$. This page puts translations and stretches and reflections together, so you can read any equation of the form $y = af\big(k(x - d)\big) + c$ and sketch it from the parent graph.

Key ideas

The four parameters

$$y = a\,f\big(k(x - d)\big) + c$$

Parameter	What it does
$a$	vertical stretch or compression by a factor of $\lvert a \rvert$; reflection in the $x$-axis if $a \lt 0$
$k$	horizontal stretch or compression by a factor of $\dfrac{1}{\lvert k \rvert}$; reflection in the $y$-axis if $k \lt 0$
$d$	horizontal translation: right if $d \gt 0$, left if $d \lt 0$
$c$	vertical translation: up if $c \gt 0$, down if $c \lt 0$

The mapping rule

Every point on the parent graph moves like this:

$$(x, y) \to \left(\frac{x}{k} + d,\ ay + c\right)$$

Notice the order inside each coordinate: stretch or reflect first, then translate. Divide $x$ by $k$, then add $d$. Multiply $y$ by $a$, then add $c$.

Factor out k first

The rule only works when the equation is written as $k(x - d)$. If it isn’t, factor:

$$y = \sqrt{2x - 6} = \sqrt{2(x - 3)}$$

Now you can see $k = 2$ and $d = 3$. Reading "$d = 6$" from $2x - 6$ is a very common mistake.

Steps for sketching

1. Write the equation in the form $y = af\big(k(x - d)\big) + c$, and read off $a$, $k$, $d$, $c$.
2. Write the mapping rule.
3. Apply it to the parent function’s key points (and asymptotes, if any).
4. Plot the new points and join them with the parent’s shape.
5. State the domain and range from the transformed graph.

On the SAT

Sliders make transformations easy to see. Define the parent, like f(x) = x^2, then type y = a f(k(x - d)) + c, and Desmos offers sliders for $a$, $k$, $d$ and $c$; drag each one to watch what it does. On the SAT, a typical question shows a transformed graph and asks which equation matches it. Often the fastest check is one key point (like the vertex) pushed through the mapping rule, and you can confirm a choice by graphing it on top of the given graph. Watch the sign of $d$: $y = f(x - 2)$ shifts right $2$, not left. See using Desmos on the SAT.

Worked examples

Example 1: Reading a quadratic

Describe the transformations of $y = x^2$ in $y = -3(x + 2)^2 + 5$. Give the vertex, the direction of opening, and the range.

Solution. $a = -3$, $k = 1$, $d = -2$, $c = 5$.

• Vertical stretch by a factor of $3$ and reflection in the $x$-axis.
• Translation $2$ units left and $5$ units up.

The vertex moves from $(0, 0)$ to $(-2, 5)$. Since $a \lt 0$, the parabola opens down, so $5$ is its highest value.

Domain $\{x \in \mathbb{R}\}$, range $\{y \in \mathbb{R} \mid y \le 5\}$.

Example 2: Sketching a square root function

Sketch $y = -2\sqrt{x + 3} + 4$, and state its domain and range.

Solution. $a = -2$, $k = 1$, $d = -3$, $c = 4$. The mapping rule is $(x, y) \to (x - 3,\ -2y + 4)$.

$y = \sqrt{x}$	$(0, 0)$	$(1, 1)$	$(4, 2)$	$(9, 3)$
$y = -2\sqrt{x + 3} + 4$	$(-3, 4)$	$(-2, 2)$	$(1, 0)$	$(6, -2)$

For example, $(4, 2) \to (4 - 3,\ -2(2) + 4) = (1, 0)$.

Starting point $(-3, 4)$; the reflection makes the graph go down to the right.

Domain $\{x \in \mathbb{R} \mid x \ge -3\}$, range $\{y \in \mathbb{R} \mid y \le 4\}$.

Example 3: Factoring out k

Describe the transformations in $y = \sqrt{2x - 6} + 1$, map the key points of $y = \sqrt{x}$, and state the domain and range.

Solution. Factor first: $y = \sqrt{2(x - 3)} + 1$. So $a = 1$, $k = 2$, $d = 3$, $c = 1$.

• Horizontal compression by a factor of $\dfrac{1}{2}$.
• Translation $3$ units right and $1$ unit up.

The mapping rule is $(x, y) \to \left(\tfrac{x}{2} + 3,\ y + 1\right)$:

$$(0, 0) \to (3, 1), \quad (1, 1) \to \left(\tfrac{7}{2}, 2\right), \quad (4, 2) \to (5, 3), \quad (9, 3) \to \left(\tfrac{15}{2}, 4\right)$$

Check one: at $x = 5$, $\sqrt{2(5) - 6} + 1 = \sqrt{4} + 1 = 3$. ✓

Domain $\{x \in \mathbb{R} \mid x \ge 3\}$, range $\{y \in \mathbb{R} \mid y \ge 1\}$.

Example 4: A reciprocal function

For $y = \dfrac{2}{x - 3} + 1$, find the asymptotes, the domain and range, and the images of $(1, 1)$, $(-1, -1)$, $\left(2, \tfrac{1}{2}\right)$, and $\left(\tfrac{1}{2}, 2\right)$.

Solution. Write it as $y = 2 \cdot \dfrac{1}{x - 3} + 1$: $a = 2$, $k = 1$, $d = 3$, $c = 1$. The mapping rule is $(x, y) \to (x + 3,\ 2y + 1)$.

• Asymptotes: $x = 0$ moves to $x = 3$, and $y = 0$ moves to $y = 2(0) + 1 = 1$.
• Domain $\{x \in \mathbb{R} \mid x \ne 3\}$, range $\{y \in \mathbb{R} \mid y \ne 1\}$.
• Points: $(1, 1) \to (4, 3)$, $(-1, -1) \to (2, -1)$, $\left(2, \tfrac{1}{2}\right) \to (5, 2)$, and $\left(\tfrac{1}{2}, 2\right) \to \left(\tfrac{7}{2}, 5\right)$.

Check: at $x = 4$, $\dfrac{2}{4 - 3} + 1 = 3$. ✓

Common mistakes

Translating before stretching. For $y$-values, multiply by $a$ first, then add $c$. Adding first gives the wrong point: for $y = 2f(x) + 3$, the point $(1, 1)$ goes to $(1, 5)$, not $(1, 8)$.

Not factoring out $k$. In $y = \sqrt{2x - 6}$, the shift is $3$ right, not $6$. Always rewrite as $k(x - d)$ first.

Getting the sign of $d$ wrong. $y = f\big(k(x + 4)\big)$ means $d = -4$, a shift left.

Multiplying $x$ by $k$ instead of dividing. In the mapping rule, the $x$-coordinate becomes $\dfrac{x}{k} + d$.

Forgetting what a reflection does to the range. In Example 2, the graph goes down from its starting point, so the range is $y \le 4$, not $y \ge 4$.

Practice

1. (Warm-up) State $a$, $k$, $d$, and $c$ for $y = 4f\big(3(x - 2)\big) - 7$.

Solution

$a = 4$, $k = 3$, $d = 2$, $c = -7$.

2. (Warm-up) Describe the transformations of $y = x^2$ in $y = 0.5(x - 4)^2 - 1$, and give the vertex.

Solution

Vertical compression by a factor of $0.5$, then $4$ units right and $1$ unit down. Vertex $(4, -1)$.

3. (Warm-up) The point $(2, 5)$ is on $y = f(x)$. Find its image on $y = 3f(x - 1) + 2$.

Solution

The rule is $(x, y) \to (x + 1,\ 3y + 2)$, so $(2, 5) \to (3, 17)$.

4. (Core) For $g(x) = 2\sqrt{x - 1} - 3$, map the key points of $y = \sqrt{x}$ and state the domain and range.

Solution

The rule is $(x, y) \to (x + 1,\ 2y - 3)$:

$$(0, 0) \to (1, -3), \quad (1, 1) \to (2, -1), \quad (4, 2) \to (5, 1), \quad (9, 3) \to (10, 3)$$

Domain $\{x \in \mathbb{R} \mid x \ge 1\}$, range $\{y \in \mathbb{R} \mid y \ge -3\}$.

5. (Core) For $h(x) = -\tfrac{1}{2}(x + 4)^2 + 6$, give the vertex, map the points $(1, 1)$, $(-1, 1)$, $(2, 4)$, and $(-2, 4)$ of $y = x^2$, and state the range.

Solution

The rule is $(x, y) \to \left(x - 4,\ -\tfrac{1}{2}y + 6\right)$. The vertex is $(-4, 6)$.

$$(1, 1) \to \left(-3, \tfrac{11}{2}\right), \quad (-1, 1) \to \left(-5, \tfrac{11}{2}\right), \quad (2, 4) \to (-2, 4), \quad (-2, 4) \to (-6, 4)$$

The parabola opens down, so the range is $\{y \in \mathbb{R} \mid y \le 6\}$.

6. (Core) Describe the transformations in $y = \sqrt{3x + 12}$, and state the domain.

Solution

Factor: $y = \sqrt{3(x + 4)}$, so $k = 3$ and $d = -4$.

Horizontal compression by a factor of $\tfrac{1}{3}$, then $4$ units left. The starting point $(0, 0)$ moves to $(-4, 0)$.

Domain $\{x \in \mathbb{R} \mid x \ge -4\}$.

Check: $(9, 3) \to \left(\tfrac{9}{3} - 4,\ 3\right) = (-1, 3)$, and $\sqrt{3(-1) + 12} = \sqrt{9} = 3$. ✓

7. (Core) The graph of $y = \dfrac{1}{x}$ is stretched vertically by a factor of $3$, reflected in the $x$-axis, and translated $2$ units right and $4$ units up. Write its equation and state its asymptotes.

Solution

$a = -3$, $d = 2$, $c = 4$:

$$y = -\frac{3}{x - 2} + 4$$

Asymptotes $x = 2$ and $y = 4$.

8. (Challenge) The function $y = f(x)$ has domain $\{x \in \mathbb{R} \mid -3 \le x \le 3\}$ and range $\{y \in \mathbb{R} \mid 0 \le y \le 4\}$. State the domain and range of $y = 2f\big(0.5(x + 1)\big) - 3$.

Solution

$a = 2$, $k = 0.5$, $d = -1$, $c = -3$. The rule is $(x, y) \to (2x - 1,\ 2y - 3)$.

Domain: $2(-3) - 1 = -7$ and $2(3) - 1 = 5$, so $\{x \in \mathbb{R} \mid -7 \le x \le 5\}$.

Range: $2(0) - 3 = -3$ and $2(4) - 3 = 5$, so $\{y \in \mathbb{R} \mid -3 \le y \le 5\}$.

9. (Challenge) The point $(6, -1)$ is on the graph of $y = -2f\big(3(x - 1)\big) + 5$. Which point on $y = f(x)$ did it come from?

Solution

The rule is $(x, y) \to \left(\tfrac{x}{3} + 1,\ -2y + 5\right)$. Work backwards from $(6, -1)$:

$$\frac{x}{3} + 1 = 6 \quad\Rightarrow\quad x = 15$$$$-2y + 5 = -1 \quad\Rightarrow\quad y = 3$$

So it came from $(15, 3)$. Check: $(15, 3) \to (5 + 1,\ -6 + 5) = (6, -1)$. ✓