Stretches, Compressions, and Reflections

Translations slide a graph without changing its shape. Stretches and compressions change its shape, pulling it taller, flatter, wider, or narrower, and reflections flip it over an axis. Together they explain the roles of $a$ and $k$ in $y = af(kx)$.

Key ideas

Vertical stretches and compressions

In $y = af(x)$, multiplying the output by $a$ multiplies every $y$-coordinate by $a$:

$$(x, y) \to (x,\ ay)$$

• $|a| \gt 1$: vertical stretch by a factor of $|a|$ (the graph gets taller).
• $0 \lt |a| \lt 1$: vertical compression by a factor of $|a|$ (the graph gets flatter).
• $a \lt 0$: also a reflection in the $x$-axis (the graph flips upside down).

Horizontal stretches and compressions

In $y = f(kx)$, multiplying the input by $k$ divides every $x$-coordinate by $k$:

$$(x, y) \to \left(\frac{x}{k},\ y\right)$$

• $|k| \gt 1$: horizontal compression by a factor of $\dfrac{1}{|k|}$ (the graph gets narrower).
• $0 \lt |k| \lt 1$: horizontal stretch by a factor of $\dfrac{1}{|k|}$ (the graph gets wider).
• $k \lt 0$: also a reflection in the $y$-axis (the graph flips left to right).

The horizontal factor is the reciprocal of $k$. For example, $y = f(2x)$ is a horizontal compression by a factor of $\dfrac{1}{2}$, because the input reaches each value twice as fast.

Pictures

Vertical changes move points up or down; horizontal changes move them left or right.

Points that don’t move

• A vertical stretch, compression, or reflection doesn’t move points on the $x$-axis (where $y = 0$).
• A horizontal stretch, compression, or reflection doesn’t move points on the $y$-axis (where $x = 0$).

These are called invariant points. They’re useful checks when you sketch.

Worked examples

Example 1: Vertical stretches and reflections

Describe each graph compared with $y = x^2$, and find the images of $(1, 1)$ and $(2, 4)$.

(a) $y = 3x^2$ $\qquad$ (b) $y = -\dfrac{1}{2}x^2$

Solution.

(a) $a = 3$: a vertical stretch by a factor of $3$. Rule $(x, y) \to (x,\ 3y)$:

$$(1, 1) \to (1, 3), \qquad (2, 4) \to (2, 12)$$

(b) $a = -\dfrac{1}{2}$: a vertical compression by a factor of $\dfrac{1}{2}$ and a reflection in the $x$-axis. Rule $(x, y) \to \left(x,\ -\tfrac{1}{2}y\right)$:

$$(1, 1) \to \left(1, -\tfrac{1}{2}\right), \qquad (2, 4) \to (2, -2)$$

The parabola is flatter and opens downward.

Example 2: Horizontal compressions and reflections

Describe each graph compared with $y = \sqrt{x}$, and state its domain.

(a) $y = \sqrt{2x}$ $\qquad$ (b) $y = \sqrt{-x}$

Solution.

(a) $k = 2$: a horizontal compression by a factor of $\dfrac{1}{2}$. Rule $(x, y) \to \left(\tfrac{x}{2},\ y\right)$, so $(4, 2) \to (2, 2)$ and $(9, 3) \to (4.5, 3)$. Check: $\sqrt{2(2)} = 2$. ✓ The domain is still $\{x \in \mathbb{R} \mid x \ge 0\}$.

(b) $k = -1$: a reflection in the $y$-axis. Rule $(x, y) \to (-x,\ y)$, so $(4, 2) \to (-4, 2)$. The graph now goes to the left, with domain $\{x \in \mathbb{R} \mid x \le 0\}$. Check: $\sqrt{-(-4)} = \sqrt{4} = 2$. ✓

Example 3: Using a mapping rule

The points $(3, 4)$ and $(-6, 1)$ are on $y = f(x)$. Find their images on $y = -2f\!\left(\tfrac{1}{3}x\right)$.

Solution. Here $a = -2$ and $k = \tfrac{1}{3}$. Dividing by $\tfrac{1}{3}$ is the same as multiplying by $3$, so the rule is:

$$(x, y) \to (3x,\ -2y)$$ $$(3, 4) \to (9, -8), \qquad (-6, 1) \to (-18, -2)$$

Example 4: A reciprocal function

Describe $y = \dfrac{4}{x}$ compared with $y = \dfrac{1}{x}$, and find the images of $(1, 1)$ and $\left(2, \tfrac{1}{2}\right)$.

Solution. $\dfrac{4}{x} = 4 \cdot \dfrac{1}{x}$, so $a = 4$: a vertical stretch by a factor of $4$. Rule $(x, y) \to (x,\ 4y)$:

$$(1, 1) \to (1, 4), \qquad \left(2, \tfrac{1}{2}\right) \to (2, 2)$$

The asymptotes stay at $x = 0$ and $y = 0$: stretching away from the $x$-axis doesn’t move either of them.

Common mistakes

Using $k$ as the horizontal factor. $y = f(3x)$ is a horizontal compression by a factor of $\dfrac{1}{3}$, not a stretch by $3$. Divide the $x$-coordinates by $k$.

Reflecting in the wrong axis. A negative $a$ (outside) flips the graph over the $x$-axis, upside down. A negative $k$ (inside) flips it over the $y$-axis, left to right.

Applying $a$ to the $x$-coordinates. $a$ only changes $y$-values, and $k$ only changes $x$-values. Write the mapping rule first.

Calling $y = \tfrac{1}{2}f(x)$ a stretch. When $0 \lt |a| \lt 1$, the graph gets flatter, so it’s a vertical compression by a factor of $\tfrac{1}{2}$.

Forgetting what a reflection does to the domain or range. $y = -\sqrt{x}$ has range $y \le 0$, and $y = \sqrt{-x}$ has domain $x \le 0$.

Practice

1. (Warm-up) Describe each transformation of $y = f(x)$.

• (a) $y = 5f(x)$
• (b) $y = \tfrac{1}{4}f(x)$
• (c) $y = -f(x)$

Solution

(a) Vertical stretch by a factor of $5$.

(b) Vertical compression by a factor of $\tfrac{1}{4}$.

(c) Reflection in the $x$-axis.

2. (Warm-up) Describe each transformation of $y = f(x)$.

• (a) $y = f(3x)$
• (b) $y = f\!\left(\tfrac{1}{2}x\right)$
• (c) $y = f(-x)$

Solution

(a) Horizontal compression by a factor of $\tfrac{1}{3}$.

(b) Horizontal stretch by a factor of $2$.

(c) Reflection in the $y$-axis.

3. (Warm-up) The point $(2, 6)$ is on $y = f(x)$. Find its image on each graph.

• (a) $y = -f(x)$
• (b) $y = f(2x)$
• (c) $y = 0.5f(x)$

Solution

(a) $(2, -6)$.

(b) Divide $x$ by $2$: $(1, 6)$.

(c) Multiply $y$ by $0.5$: $(2, 3)$.

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

Solution

Vertical stretch by a factor of $3$ and reflection in the $x$-axis. Rule $(x, y) \to (x,\ -3y)$:

$$(0, 0) \to (0, 0), \quad (1, 1) \to (1, -3), \quad (4, 2) \to (4, -6), \quad (9, 3) \to (9, -9)$$

Domain $\{x \in \mathbb{R} \mid x \ge 0\}$, range $\{y \in \mathbb{R} \mid y \le 0\}$.

5. (Core) For $h(x) = \sqrt{-4x}$, describe the transformations, map the key points of $y = \sqrt{x}$, and state the domain and range.

Solution

$k = -4$: horizontal compression by a factor of $\tfrac{1}{4}$ and reflection in the $y$-axis. Rule $(x, y) \to \left(-\tfrac{x}{4},\ y\right)$:

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

Check: $\sqrt{-4(-1)} = \sqrt{4} = 2$. ✓

Domain $\{x \in \mathbb{R} \mid x \le 0\}$, range $\{y \in \mathbb{R} \mid y \ge 0\}$.

6. (Core) Write the equation of $y = x^2$ after a vertical compression by a factor of $\tfrac{1}{3}$ and a reflection in the $x$-axis.

Solution

$$y = -\tfrac{1}{3}x^2$$

7. (Core) Show that $y = \dfrac{1}{2x}$, a horizontal compression of $y = \dfrac{1}{x}$, is also a vertical compression of it. Check with the point $(1, 1)$.

Solution

$$\frac{1}{2x} = \frac{1}{2} \cdot \frac{1}{x}$$

So it’s also a vertical compression by a factor of $\tfrac{1}{2}$.

As a horizontal compression, $(1, 1) \to \left(\tfrac{1}{2}, 1\right)$. Check it’s on the graph: $\dfrac{1}{2 \cdot \frac{1}{2}} = 1$. ✓

8. (Challenge) Show that compressing $y = x^2$ horizontally by a factor of $\tfrac{1}{2}$ gives the same graph as stretching it vertically by a factor of $4$.

Solution

The horizontal compression is $y = (2x)^2$, and

$$(2x)^2 = 4x^2$$

which is the vertical stretch by a factor of $4$.

Check with a point: the compression sends $(1, 1)$ to $\left(\tfrac{1}{2}, 1\right)$, and on $y = 4x^2$, $4\left(\tfrac{1}{2}\right)^2 = 1$. ✓

9. (Challenge) The function $y = f(x)$ has domain $\{x \in \mathbb{R} \mid -2 \le x \le 6\}$ and range $\{y \in \mathbb{R} \mid -1 \le y \le 3\}$. State the domain and range of $y = -2f(0.5x)$.

Solution

The rule is $(x, y) \to (2x,\ -2y)$.

Domain: double the endpoints, $\{x \in \mathbb{R} \mid -4 \le x \le 12\}$.

Range: multiply the endpoints by $-2$ to get $2$ and $-6$. The negative factor swaps which end is bigger, so the range is $\{y \in \mathbb{R} \mid -6 \le y \le 2\}$.