Transformations of Exponential Functions

Every exponential graph you’ll meet is a stretched, flipped, or shifted copy of a parent like $y = 2^x$. The same transformation rules you used for other functions apply here. The one new thing to watch is the horizontal asymptote, which moves up and down with the graph.

Key ideas

The transformed exponential function

$$y = a \cdot b^{k(x - d)} + c$$

The parameters do the same jobs as before:

Parameter	Effect
$a$	vertical stretch or compression by $\lvert a \rvert$; reflection in the $x$-axis if $a \lt 0$
$k$	horizontal stretch or compression by $\dfrac{1}{\lvert k \rvert}$; reflection in the $y$-axis if $k \lt 0$
$d$	horizontal translation
$c$	vertical translation

The mapping rule is $(x, y) \to \left(\dfrac{x}{k} + d,\ ay + c\right)$.

The asymptote moves with c

The parent $y = b^x$ has asymptote $y = 0$. After the transformation:

• the horizontal asymptote is $y = c$
• the range is $y \gt c$ if $a \gt 0$, or $y \lt c$ if $a \lt 0$
• the domain is still all real numbers

Horizontal changes ($k$ and $d$) don’t move a horizontal line, and stretching $y = 0$ leaves it at $0$. Only $c$ moves the asymptote.

Key points to map

For $y = b^x$, use $\left(-1, \tfrac{1}{b}\right)$, $(0, 1)$, $(1, b)$, and $(2, b^2)$, together with the asymptote.

Worked examples

Example 1: Describing transformations

Describe how $y = 3(2)^{x - 4} + 1$ relates to $y = 2^x$, and state the asymptote and range.

Solution. $a = 3$, $k = 1$, $d = 4$, $c = 1$.

• Vertical stretch by a factor of $3$.
• Translation $4$ units right and $1$ unit up.

Asymptote $y = 1$. Since $a \gt 0$, the range is $\{y \in \mathbb{R} \mid y \gt 1\}$.

Example 2: Sketching

Sketch $y = 2^{x + 1} - 3$, and state the asymptote, $y$-intercept, domain, and range.

Solution. $d = -1$ and $c = -3$, so the mapping rule is $(x, y) \to (x - 1,\ y - 3)$.

$y = 2^x$	$\left(-1, \tfrac{1}{2}\right)$	$(0, 1)$	$(1, 2)$	$(2, 4)$	$(3, 8)$
$y = 2^{x + 1} - 3$	$\left(-2, -\tfrac{5}{2}\right)$	$(-1, -2)$	$(0, -1)$	$(1, 1)$	$(2, 5)$

The asymptote moves down with the graph, from $y = 0$ to $y = -3$.

Asymptote $y = -3$, $y$-intercept $-1$ (check: $2^1 - 3 = -1$ ✓), domain $\{x \in \mathbb{R}\}$, range $\{y \in \mathbb{R} \mid y \gt -3\}$.

Example 3: A reflection

For $y = -2(3)^{x - 1} + 4$, map the points $\left(-1, \tfrac{1}{3}\right)$, $(0, 1)$, and $(1, 3)$ of $y = 3^x$, and state the asymptote and range.

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

$$\left(-1, \tfrac{1}{3}\right) \to \left(0, \tfrac{10}{3}\right), \qquad (0, 1) \to (1, 2), \qquad (1, 3) \to (2, -2)$$

Asymptote $y = 4$. Because $a \lt 0$, the graph is below the asymptote: the range is $\{y \in \mathbb{R} \mid y \lt 4\}$.

Example 4: Writing the equation from properties

An exponential function of the form $y = a(2)^x + c$ has asymptote $y = 2$ and passes through $(0, 5)$. Find its equation.

Solution. The asymptote gives $c = 2$. Substitute $(0, 5)$:

$$5 = a(2)^0 + 2 = a + 2 \quad\Rightarrow\quad a = 3$$

So $y = 3(2)^x + 2$. Check another point: at $x = 1$, $y = 3(2) + 2 = 8$, so $(1, 8)$ should be on the graph.

Common mistakes

Moving the asymptote sideways. The asymptote is horizontal, so $d$ doesn’t affect it. Only $c$ does: $y = 2^{x + 1} - 3$ has asymptote $y = -3$.

Getting the range wrong after a reflection. If $a \lt 0$, the graph is below the asymptote, so the range uses $\lt$.

Reading the horizontal shift with the wrong sign. $2^{x + 1}$ moves the graph left $1$.

Not factoring out $k$. $y = 2^{2x - 4}$ is $2^{2(x - 2)}$: a compression by $\tfrac{1}{2}$ and a shift of $2$ right, not $4$.

Forgetting that the $y$-intercept changes. It’s no longer $1$. Substitute $x = 0$ to find it.

Practice

1. (Warm-up) State the asymptote and range of $y = 5^x - 2$.

Solution

Asymptote $y = -2$; range $\{y \in \mathbb{R} \mid y \gt -2\}$.

2. (Warm-up) Describe the transformation that takes $y = 4^x$ to $y = 4^{x + 3}$.

Solution

A translation $3$ units left.

3. (Warm-up) Find the $y$-intercept of $y = 3(2)^x + 1$.

Solution

$y = 3(2)^0 + 1 = 3 + 1 = 4$.

4. (Core) For $y = 3^{x - 2} + 1$, map the points $\left(-1, \tfrac{1}{3}\right)$, $(0, 1)$, $(1, 3)$, and $(2, 9)$, and state the asymptote and range.

Solution

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

$$\left(-1, \tfrac{1}{3}\right) \to \left(1, \tfrac{4}{3}\right), \quad (0, 1) \to (2, 2), \quad (1, 3) \to (3, 4), \quad (2, 9) \to (4, 10)$$

Asymptote $y = 1$; range $\{y \in \mathbb{R} \mid y \gt 1\}$.

5. (Core) For $y = -\left(\tfrac{1}{2}\right)^x + 3$, describe the transformations, find the $y$-intercept, and state the range.

Solution

A reflection in the $x$-axis, then a translation $3$ units up.

$y$-intercept: $-1 + 3 = 2$.

The asymptote is $y = 3$, and the graph is below it, so the range is $\{y \in \mathbb{R} \mid y \lt 3\}$.

6. (Core) Describe the transformations in $y = 2^{2x - 6}$, and map the points $(0, 1)$, $(1, 2)$, and $(2, 4)$ of $y = 2^x$.

Solution

Factor: $y = 2^{2(x - 3)}$. A horizontal compression by a factor of $\tfrac{1}{2}$, then a translation $3$ units right. The rule is $(x, y) \to \left(\tfrac{x}{2} + 3,\ y\right)$:

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

Check: $2^{2(4) - 6} = 2^2 = 4$. ✓

7. (Core) A function of the form $y = a(3)^x + c$ has asymptote $y = -1$ and passes through $(0, 3)$. Find its equation, and check that it passes through $(1, 11)$.

Solution

$c = -1$, and $3 = a + (-1)$ gives $a = 4$. So $y = 4(3)^x - 1$.

At $x = 1$: $4(3) - 1 = 11$. ✓

8. (Challenge) Show that $y = 9^x$ is a horizontal compression of $y = 3^x$, and that $y = 3^{x + 2}$ is a vertical stretch of $y = 3^x$. Give the factor each time.

Solution

$9^x = (3^2)^x = 3^{2x}$: a horizontal compression by a factor of $\tfrac{1}{2}$.

$3^{x + 2} = 3^x \times 3^2 = 9(3^x)$: a vertical stretch by a factor of $9$. (It’s also a translation $2$ units left: for exponential functions, both descriptions give the same graph.)

9. (Challenge) The point $(3, 13)$ is on the graph of $y = 3(2)^{x - 1} + 1$. Which point on $y = 2^x$ did it come from?

Solution

The rule is $(x, y) \to (x + 1,\ 3y + 1)$. Work backwards:

$$x + 1 = 3 \Rightarrow x = 2, \qquad 3y + 1 = 13 \Rightarrow y = 4$$

It came from $(2, 4)$, which is on $y = 2^x$ since $2^2 = 4$. ✓