Exponential functions

So far, expressions like $x^2$ or $x^5$ have had the variable in the base and a fixed number in the exponent. An exponential function flips this: the base is fixed and the variable sits in the exponent.

Because the input is itself the exponent, the exponent rules we already know tell us exactly what $f$ does at familiar values:

• $a^0 = 1$, so $f(0) = 1$ for every base.
• $a^{-n} = \dfrac{1}{a^n}$, so negative inputs give reciprocals.
• $a^{m/n} = \left(\sqrt[n]{a}\right)^m$, so fractional inputs involve roots.

Let $f(x) = 3^x$. Evaluate $f(0)$, $f(-2)$, and $f\!\left(\dfrac{3}{2}\right)$.

Solution

First, $f(0)$. Any non-zero base raised to the power $0$ equals $1$:

Next, $f(-2)$. A negative exponent gives the reciprocal:

Finally, $f\!\left(\dfrac{3}{2}\right)$. A fractional exponent combines a root and a power. The denominator $2$ gives the square root and the numerator $3$ gives the power. Taking the root first:

So the three outputs are $f(0) = 1$, $f(-2) = \dfrac{1}{9}$, and $f\!\left(\dfrac{3}{2}\right) = 3\sqrt{3}$.

Whether an exponential function rises or falls depends entirely on its base.

The key idea is what happens when $x$ increases by $1$. Since $a^{x+1} = a \cdot a^x$, each unit step in $x$ multiplies the output by the base $a$. When $a > 1$, multiplying by $a$ makes the output larger; when $0 < a < 1$, multiplying by $a$ makes it smaller.

Growth and decay are two sides of the same coin. Notice that

so a decay curve with base $\dfrac{1}{a}$ is the growth curve $a^x$ reflected across the $y$-axis. For every $a > 1$, the two curves are mirror images of each other:

Every decay function can be rewritten as a growth function with a negated exponent, and vice versa.

Classify $f(x) = 5^x$ and $g(x) = \left(\dfrac{2}{5}\right)^x$ as growth or decay. Then show that $g(x)$ can be rewritten using base $\dfrac{5}{2}$.

Solution

$f(x) = 5^x$: The base is $5$. Since $5 > 1$, each unit increase in $x$ multiplies the output by $5$, so $f$ represents exponential growth.

$g(x) = \left(\dfrac{2}{5}\right)^x$: The base is $\dfrac{2}{5}$. Since $0 < \dfrac{2}{5} < 1$, each unit increase in $x$ multiplies the output by $\dfrac{2}{5}$, making it smaller. So $g$ represents exponential decay.

Rewriting $g$: We can express $\dfrac{2}{5}$ as the reciprocal of $\dfrac{5}{2}$:

Substituting into $g(x)$ and applying the power-of-a-power exponent rule:

This confirms the relationship: decay with base $\dfrac{2}{5}$ is the same as growth with base $\dfrac{5}{2}$ reflected across the $y$-axis.

Every exponential function $f(x) = a^x$ shares a core set of graphical features, no matter what the base is.

The base determines the shape:

• When $a > 1$ (growth), the curve rises steeply to the right and flattens towards zero on the left.
• When $0 < a < 1$ (decay), the curve falls from left to right, the mirror image of a growth curve.
• A larger base produces steeper growth. For example, $3^x$ rises faster than $2^x$.

Use the slider below to adjust the base and watch how the curve changes. As $a$ crosses $1$, the graph flips from decay to growth.

For $f(x) = 2^x$, identify the $y$-intercept and horizontal asymptote, and explain why the graph is always positive. Then sketch $g(x) = \left(\dfrac{1}{2}\right)^x$ on the same axes and describe the relationship between the two graphs.

Solution

Features of $f(x) = 2^x$:

• $y$-intercept: Setting $x = 0$ gives $f(0) = 2^0 = 1$, so the $y$-intercept is $(0, 1)$.
• Horizontal asymptote: As $x$ becomes very negative, $2^x$ gets arbitrarily close to $0$ but never reaches it. The horizontal asymptote is $y = 0$.
• Always positive: Since $2^x > 0$ for all $x$, the curve stays above the $x$-axis and never crosses it - there is no $x$-intercept.

Sketching $g(x) = \left(\dfrac{1}{2}\right)^x$:

Since $\left(\dfrac{1}{2}\right)^x = 2^{-x}$, replacing $x$ with $-x$ reflects the graph across the $y$-axis. The decay curve $g$ is the mirror image of the growth curve $f$. Both curves pass through $(0, 1)$ and share the same asymptote $y = 0$.

The blue growth curve $f(x) = 2^x$ and the red decay curve $g(x) = \left(\dfrac{1}{2}\right)^x$ are reflections of each other in the $y$-axis. They share the $y$-intercept $(0, 1)$ and the asymptote $y = 0$.