Exponential functions and their graphs

An exponential function has the variable in the exponent. Its basic form is

$$f(x)=a^x,$$

where the base satisfies $a>0$ and $a\ne1$. Exponential functions describe repeated multiplication by a constant factor. They are the mathematical language of compound interest, population growth, radioactive decay and many other processes.

The distinction between a power function and an exponential function is essential:

$$\underbrace{x^3}_{\text{power function: variable base}} \qquad \underbrace{3^x}_{\text{exponential function: variable exponent}}.$$

Before you begin

You should be able to:

• use the laws of indices, including zero, negative and fractional indices;
• read coordinates, intercepts and asymptotes from a graph;
• apply graph transformations.

Why the restrictions on the base matter

We take $a>0$ so that $a^x$ is real for every real $x$. A negative base can fail to give a real value. For example, $(-2)^{1/2}$ is not real, although $(-2)^3$ is real. It therefore does not define the continuous real function required here.

We also exclude $a=1$ because $1^x=1$ for every $x$. That is a constant function, not exponential growth or decay.

The index laws extend the meaning of $a^x$ beyond positive integer values of $x$:

$$a^0=1, \qquad a^{-x}=\frac{1}{a^x}, \qquad a^{m/n}=\sqrt[n]{a^m}.$$

For example,

$$4^{-3/2}=\frac{1}{4^{3/2}} =\frac{1}{(\sqrt4)^3} =\frac18.$$

These rules explain points on the graph between integer inputs and to the left of the $y$-axis.

Growth when $\boldsymbol{a>1}$

Consider $f(x)=2^x$.

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$2^x$	$\frac18$	$\frac14$	$\frac12$	$1$	$2$	$4$	$8$

Moving one unit to the right multiplies the output by $2$. More generally,

$$\frac{a^{x+1}}{a^x}=a.$$

If $a>1$, this constant factor is greater than $1$, so the function increases. Every graph $y=a^x$ with $a>1$ has these features:

• domain $x\in\mathbb R$;
• range $y>0$;
• $y$-intercept $(0,1)$ because $a^0=1$;
• no $x$-intercept because $a^x$ is always positive;
• horizontal asymptote $y=0$;
• $a^x\to\infty$ as $x\to\infty$;
• $a^x\to0$ as $x\to-\infty$.

The graph approaches the $x$-axis but never reaches it. Saying that $a^x=0$ for a sufficiently negative $x$ is incorrect. A calculator may round a very small positive number to zero, but the exact value remains positive.

Worked example 1: use the constant factor

Suppose $f(x)=3^x$ and $f(p)=7$. Find $f(p+2)$ and $f(p-1)$ without finding $p$.

Use the index laws:

$$f(p+2)=3^{p+2}=3^p\times3^2=7\times9=63.$$

Similarly,

$$f(p-1)=3^{p-1}=\frac{3^p}{3}=\frac73.$$

The unknown input is irrelevant. A change of $2$ in the input multiplies the output by $3^2$.

Decay when $\boldsymbol{0<a<1}$

If the base lies between $0$ and $1$, increasing $x$ multiplies the output by a factor less than $1$. The function therefore decreases. For example,

$$\left(\frac12\right)^x=2^{-x}.$$

Its graph is the reflection of $y=2^x$ in the $y$-axis. It still has domain $\mathbb R$, range $y>0$, intercept $(0,1)$ and asymptote $y=0$, but its end behaviour is reversed:

$$\left(\frac12\right)^x\to0 \text{ as }x\to\infty, \qquad \left(\frac12\right)^x\to\infty \text{ as }x\to-\infty.$$

This is exponential decay. Decay means repeated multiplication by a positive factor smaller than $1$. It does not mean the output is negative.

Worked example 2: rewrite the base before sketching

Describe the graph of

$$y=9^{-x/2}.$$

Rewrite the expression using index laws:

$$9^{-x/2}=\left(9^{-1/2}\right)^x =\left(\frac13\right)^x.$$

The base is between $0$ and $1$, so the graph decreases. It passes through $(0,1)$ and has horizontal asymptote $y=0$. Useful exact points are

$$x=-1\Rightarrow y=3, \qquad x=1\Rightarrow y=\frac13.$$

The natural exponential function

The number

$$e=2.718281828\ldots$$

is the natural base for exponentials. The function $y=e^x$ has exactly the same qualitative shape as every $y=a^x$ with $a>1$: it passes through $(0,1)$, is always positive and has asymptote $y=0$.

Its special importance comes from calculus:

$$\frac{\mathrm d}{\mathrm dx}e^x=e^x.$$

At every point, the gradient of $y=e^x$ equals its height. This makes $e^x$ the simplest function for describing continuously changing quantities. The derivative property is developed in differentiating standard functions.

On a calculator, use the $e^x$ key rather than rounding $e$ first.

Worked example 3: evaluate accurately

Evaluate $5e^{-0.4}$ to three significant figures.

Enter the complete expression:

$$5e^{-0.4}=3.351600\ldots$$

Therefore

$$5e^{-0.4}=3.35 \quad \text{to three significant figures}.$$

The negative exponent makes $e^{-0.4}=1/e^{0.4}<1$. It does not make the answer negative.

Transforming exponential graphs

For

$$y=Aa^{x-h}+k,$$

start with $y=a^x$ and apply the transformations carefully.

Feature	Effect
$x-h$	translate $h$ units right
$+k$	translate $k$ units up
factor $A$	stretch parallel to the $y$-axis by scale factor $
$A<0$	also reflect in the $x$-axis

The original asymptote $y=0$ becomes $y=k$. The transformed graph has range $y>k$ when $A>0$, and $y<k$ when $A<0$.

Do not assume the $y$-intercept is still $1$. It must be found by substituting $x=0$.

Worked example 4: sketch a translated exponential

Sketch

$$y=2^{x-1}+3,$$

stating its asymptote, intercept and range.

The graph $y=2^x$ is translated $1$ unit right and $3$ units up.

The horizontal asymptote is therefore

$$y=3.$$

For the $y$-intercept, set $x=0$:

$$y=2^{-1}+3=\frac12+3=\frac72.$$

So the graph crosses the $y$-axis at $(0,7/2)$. It is increasing and remains above its asymptote, giving the range

$$y>3.$$

It has no $x$-intercept because $2^{x-1}+3$ is always greater than $3$.

Worked example 5: reflection and an intercept

Describe

$$y=4-3e^{-x}.$$

Write it as $y=-3e^{-x}+4$. The $-x$ reflects $y=e^x$ in the $y$-axis, so $e^{-x}$ decreases. Multiplication by $-3$ reflects it in the $x$-axis and stretches it. Adding $4$ moves it upwards.

As $x\to\infty$, $e^{-x}\to0$, so the horizontal asymptote is $y=4$. Since $-3e^{-x}<0$, the graph always lies below this asymptote and has range $y<4$.

Its $y$-intercept is

$$y=4-3e^0=1,$$

so it passes through $(0,1)$. For the $x$-intercept,

$$4-3e^{-x}=0 \quad\Longrightarrow\quad e^{-x}=\frac43.$$

Finding the exact value of $x$ requires a logarithm. The graph still tells us that there is one negative root because the function is increasing, is $1$ at $x=0$, and tends to $-\infty$ as $x\to-\infty$.

Finding an exponential rule from information

A common model has the form

$$y=Ab^x.$$

Here $A=y(0)$ is the initial value, while $b$ is the multiplication factor for each increase of $1$ in $x$.

Worked example 6: determine the parameters

An exponential function has the form $f(x)=Ab^x$. It passes through $(0,6)$ and $(2,54)$, where $A>0$ and $b>0$. Find $f(x)$.

Using $(0,6)$:

$$6=Ab^0=A,$$

so $A=6$. Now use $(2,54)$:

$$54=6b^2 \quad\Longrightarrow\quad b^2=9.$$

Although algebraically $b=\pm3$, an exponential base must be positive, so $b=3$. Hence

$$\boxed{f(x)=6\cdot3^x}.$$

Check: $f(2)=6\cdot9=54$.

Common misconceptions

• $a^{x+1}$ is $a\cdot a^x$, not $a^x+1$.
• $a^x+a^x=2a^x$, not $a^{2x}$. Index laws apply to multiplication and division of powers, not addition.
• A negative exponent gives a reciprocal: $e^{-x}=1/e^x$. The value is still positive.
• Exponential decay approaches zero without becoming zero.
• In $a^{x-h}$, the graph moves right by $h$. Horizontal transformations act in the opposite direction to the sign seen inside the function.
• A horizontal asymptote need not be the $x$-axis after a vertical translation.

Check your understanding

1. Features of a graph

State the domain, range, $y$-intercept and horizontal asymptote of $y=5^x$.

Answer

The domain is $x\in\mathbb R$, the range is $y>0$, the $y$-intercept is $(0,1)$ and the horizontal asymptote is $y=0$. Since $5>1$, the function is increasing.

2. Growth or decay

Without a calculator, decide whether $y=8^{-x/3}$ represents growth or decay, and simplify its base.

Answer

$$8^{-x/3}=\left(8^{-1/3}\right)^x =\left(\frac12\right)^x.$$

The base is $1/2$, so it is exponential decay.

3. A transformed graph

For $y=-2\cdot3^x+5$, find the horizontal asymptote, $y$-intercept and range.

Answer

The asymptote is $y=5$. At $x=0$,

$$y=-2\cdot3^0+5=3,$$

so the $y$-intercept is $(0,3)$. Since $-2\cdot3^x<0$, the graph lies below the asymptote and its range is $y<5$.

4. Find the function

The function $g(x)=Ab^x$ passes through $(0,10)$ and $(1,8)$. Find $A$ and $b$, and state whether $g$ shows growth or decay.

Answer

From $g(0)=10$, $A=10$. Then

$$8=10b \quad\Longrightarrow\quad b=0.8.$$

Thus $g(x)=10(0.8)^x$. Since $0<b<1$, it shows decay.

Where next?

• Learn how logarithms reverse exponentiation.
• Use logarithms to solve exponential and logarithmic equations.
• Apply these functions to exponential growth and decay models.
• Study how exponential relationships can be tested by linearising data.