Exponential Functions — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

A cup of tea cools, a savings account grows, a medicine leaves the bloodstream. Each changes by the same factor every hour or year, and each levels off toward a value it never quite reaches. The function that does this is f(x) = ab^x.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Over equal steps, outputs go $2, 6, 18, 54$ . What kind of function fits?

Exponential, ratio $3$
Linear, slope $3$
Quadratic
Linear, slope $4$

Show the answer

Equal steps, constant ratio $3$ .

Correct: Exponential, ratio $3$ :
Linear, slope $3$ :
Quadratic:
Linear, slope $4$ :

2. What does $\lim_{x \to \infty} f(x) = 2$ mean?

As $x$ grows without bound, $f(x)$ approaches $2$
$f(2) = \infty$
$f$ is never $2$
$f(x) = 2$ for large $x$

Show the answer

Limit notation describes the value the outputs approach.

Correct: As $x$ grows without bound, $f(x)$ approaches $2$ :
$f(2) = \infty$ :
$f$ is never $2$ :
$f(x) = 2$ for large $x$ :

Part 4 · See it

See it first

The increasing graph f(x) = 3(2)^x and the decreasing graph g(x) = 3(1/2)^x, both concave up, both through (0, 3), approaching the dashed asymptote y = 0 one on the left and one on the right. — Growth $f(x) = 3(2)^x$ and decay $g(x) = 3\left(\frac{1}{2}\right)^x$ : both concave up, both through $(0, 3)$ , both with the horizontal asymptote $y = 0$ , one on the left and one on the right. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$b > 1$ (growth) or $0 < b < 1$ (decay), with $a > 0$ the function is increasing or decreasing everywhere, and concave up
$a < 0$ the graph is reflected across the $x$ -axis: concave down
The output is multiplied by $b$ foreverone side approaches the asymptote: $\lim_{x \to \infty} ab^x = 0$ for decay, $\lim_{x \to -\infty} ab^x = 0$ for growth
Add $k$ : $ab^x + k$ the asymptote moves to $y = k$

Part 6 · Key ideas

Key ideas

$f(x) = ab^x$ ( $a \ne 0$ , $b > 0$ , $b \ne 1$ ) is always increasing or always decreasing, has one concavity, and has no extrema or points of inflection.
Its graph has a horizontal asymptote; write the end behavior in limit notation, such as $\lim_{x \to -\infty} 3(2)^x$ $= 0$ .
The base is the growth factor: a $4\%$ increase is $b = 1.04$ ; a $15\%$ decrease is $b = 0.85$ . The natural base $e \approx 2.718$ is a base used in many models.

Part 7 · Misconception

A common mistake

The wrong idea: $h(x) = 2\left(\frac{1}{2}\right)^x + 1$ approaches $0$ as $x \to \infty$ .

What actually happens: The $+1$ shifts the asymptote: $\lim_{x \to \infty} h(x) = 1$ . Find the limit of the exponential part, then add the shift.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $h(x) = 2\left(\frac{1}{2}\right)^x + 1$

1. Which statement describes the end behavior of $h$ as $x$ increases without bound?

$\lim_{x \to \infty} h(x) = 0$
$\lim_{x \to \infty} h(x) = 1$
$\lim_{x \to \infty} h(x) = \infty$
$\lim_{x \to 1} h(x) = \infty$

Show the answer

As $x$ grows, $\left(\frac{1}{2}\right)^x$ shrinks toward $0$ , so $h(x)$ approaches $2 \cdot 0 + 1 = 1$ : $\lim_{x \to \infty} h(x) = 1$ . The line $y = 1$ is a horizontal asymptote.

$\lim_{x \to \infty} h(x) = 0$ : This is the limit of $2(\frac{1}{2})^x$ alone; the $+1$ shifts the asymptote up.
Correct: $\lim_{x \to \infty} h(x) = 1$ : The term $2(\frac{1}{2})^x$ goes to $0$ and $1$ remains.
$\lim_{x \to \infty} h(x) = \infty$ : The graph falls as $x$ increases; it does not grow.
$\lim_{x \to 1} h(x) = \infty$ : This describes the input approaching $1$ , which is not end behavior.

2. Which statement describes $h$ as $x$ decreases without bound?

$\lim_{x \to -\infty} h(x) = 1$
$\lim_{x \to -\infty} h(x) = -\infty$
$\lim_{x \to -\infty} h(x) = \infty$
$\lim_{x \to \infty} h(x) = -\infty$

Show the answer

For negative $x$ , $\left(\frac{1}{2}\right)^x = 2^{-x}$ grows without bound: $h(-10) = 2 \cdot 1024 + 1$ . So $\lim_{x \to -\infty} h(x) = \infty$ .

$\lim_{x \to -\infty} h(x) = 1$ : The asymptote $y = 1$ is on the right side, as $x \to \infty$ .
$\lim_{x \to -\infty} h(x) = -\infty$ : The outputs are all above $1$ , so they cannot go to $-\infty$ .
Correct: $\lim_{x \to -\infty} h(x) = \infty$ : Going left, the outputs grow without bound.
$\lim_{x \to \infty} h(x) = -\infty$ : This is about the right side, and the outputs there approach $1$ .

3. Which statement about $h$ on the interval $(-3, 5)$ is true?

$h$ is decreasing, and the rate of change of $h$ is increasing
$h$ is decreasing, and the rate of change of $h$ is decreasing
$h$ is increasing, and the graph is concave up
$h$ is decreasing until $x = 0$ , then increasing

Show the answer

The outputs fall as $x$ increases, so $h$ is decreasing. The graph is concave up: the average rates of change over equal intervals are negative but closer to $0$ each time ( $h$ drops $4$ , then $2$ , then $1$ from $x = -2$ to $x = 1$ ), so the rate of change of $h$ is increasing.

Correct: $h$ is decreasing, and the rate of change of $h$ is increasing: Falling, and falling less steeply: rates increasing toward $0$ .
$h$ is decreasing, and the rate of change of $h$ is decreasing: Rates like $-4, -2, -1$ are increasing, which means concave up.
$h$ is increasing, and the graph is concave up: The base $\frac{1}{2}$ makes $h$ decrease.
$h$ is decreasing until $x = 0$ , then increasing: An exponential function has no turning point.

Table

Values of a function $k$

Selected values of the function k
$x$	0	1	2	3
$k(x)$	7	11	19	35

4. The outputs $7, 11, 19, 35$ do not have a constant ratio. Which statement shows that $k$ is still an exponential function shifted vertically?

The differences $4, 8, 16$ between outputs show that $k$ is linear
The outputs increase faster and faster, so $k$ is exponential
The values $k(x) - 3$ are $4, 8, 16, 32$ : they double each step
The ratio of the last output to the first, $\frac{35}{7} = 5$ , is constant

Show the answer

Subtract $3$ from every output: $4, 8, 16, 32$ . Over equal steps these have a constant ratio of $2$ , so $k(x) - 3 = 4(2)^x$ and $k(x) = 4(2)^x + 3$ , an exponential function shifted up $3$ .

The differences $4, 8, 16$ between outputs show that $k$ is linear: Linear needs constant differences; these double.
The outputs increase faster and faster, so $k$ is exponential: Increasing outputs fit many functions.
Correct: The values $k(x) - 3$ are $4, 8, 16, 32$ : they double each step: Shifted outputs that are proportional.
The ratio of the last output to the first, $\frac{35}{7} = 5$ , is constant: One ratio across the table does not show a pattern over equal steps.

5. A car loses $15\%$ of its value each year. Which model gives its value $V$ after $t$ years, if it cost $$20{,}000$ ?

$V(t) = 20{,}000(0.15)^t$
$V(t) = 20{,}000(1.15)^t$
$V(t) = 20{,}000 - 0.15t$
$V(t) = 20{,}000(0.85)^t$

Show the answer

Losing $15\%$ leaves $85\%$ , so the decay factor is $1 - 0.15 = 0.85$ : $V(t) = 20{,}000(0.85)^t$ .

$V(t) = 20{,}000(0.15)^t$ : This keeps only $15\%$ each year.
$V(t) = 20{,}000(1.15)^t$ : This grows by $15\%$ .
$V(t) = 20{,}000 - 0.15t$ : This subtracts $0.15$ dollars a year.
Correct: $V(t) = 20{,}000(0.85)^t$ : Keep $85\%$ each year.

6. Why does an exponential function $f(x) = ab^x$ have no local maximum or minimum?

Its outputs are all positive, so the graph stays above the $x$ -axis
Its graph has a horizontal asymptote that it never reaches
It is concave up, so the graph bends the same way everywhere
Each unit step multiplies the output by $b$ , so it moves one way only

Show the answer

Over every unit step the output is multiplied by $b$ . If $b > 1$ every step moves the output away from $0$ in the same direction; if $0 < b < 1$ , every step moves it toward $0$ . The function is increasing everywhere or decreasing everywhere, so it never turns.

Its outputs are all positive, so the graph stays above the $x$ -axis: If $a < 0$ the outputs are negative, and there is still no extremum.
Its graph has a horizontal asymptote that it never reaches: Many functions with asymptotes have extrema.
It is concave up, so the graph bends the same way everywhere: Concave-up functions such as $x^2$ can have a minimum.
Correct: Each unit step multiplies the output by $b$ , so it moves one way only: One fixed factor, one direction.

Part 9 · Summary

Summary

f(x) = ab^x

is monotonic with one concavity and a horizontal asymptote, written in limit notation. The base is the growth factor (

1 +

the percent change as a decimal). Adding

k

moves the asymptote to

y = k

; the natural base

e

is one choice of

b

.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of h(x)=2(12)x+1h(x) = 2\left(\frac{1}{2}\right)^x + 1

Values of a function kk

Summary

What comes next

Connections

The graph of $h(x) = 2\left(\frac{1}{2}\right)^x + 1$

Values of a function $k$