Exponential Functions

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characteristics

The formula template for the basic exponential function looks like

$a \, r^x \, \text { with } \, a, r \in \mathbb {R} \, | \, r > 0$

As we have seen before, the coefficient $a$ controls vertical stretching or compression. The sign of $a$ dictates the sign of our function values. $r$ dictates a growing or decaying function.

We have four combinations

$a>0$ and $r>1$
$a>0$ and $r<1$
$a<0$ and $r>1$
$a<0$ and $r<1$

For graphs of basic exponential functions, the horizontal axis is a horizontal asymptote in one direction or the other.

When $r > 1$ we have a growing function.

$\lim _{x \to -\infty } a \, r^x = 0 \, \text { and } \, \lim _{x \to \infty } a \, r^x = \pm \infty$

The sign of $a$ dictates if the unbounded growth is positive or negative.

When $r<1$ , the horizontal axis is still a horizontal asymptote, just in the other direction. The function now decays.

$\lim _{x \to -\infty } a \, r^x = \pm \infty \, \text { and } \, \lim _{x \to \infty } a \, r^x = 0$

The sign of $a$ dictates if the function decays through positive or negative values.

All four graphs share a common structure.

All have the horizontal axis as an asymptote.
All are a distance of $1$ from the asymptote, when the exponent equals $0$ .

These are the important aspects or characteristics that we use when shifting and stretching graphs of exponential functions.

Exponential Behavior

Basic exponential functions, $a \cdot r^x$ , are either increasing functions or decreasing functions.

$\blacktriangleright$ Base Greater than $1$ : $r > 1$

greater positive exponents mean multiplying by the base more, which results in larger values.
greater negative exponents mean multiplying by the reciprocal of the base more, which results in smaller values.

The coefficient in front, $a$ , tells us if this larger/smaller value is larger positively or negatively.

$a > 0$ and $r > 1$ : increasing function
$a < 0$ and $r > 1$ : decreasing function

$\blacktriangleright$ Base Less than $1$ : $r < 1$

greater positive exponents mean multiplying by the base more, which results in smaller values.
greater negative exponents mean multiplying by the reciprocal of the base more, which results in larger values.

The coefficient in front, $a$ , tells us if this larger/smaller value is larger positively or negatively.

$a > 0$ and $r < 1$ : decreasing function
$a < 0$ and $r < 1$ : increasing function

Exponential Function

Analyze $f(x) = \frac {1}{3} \, 2^{x+5} - 7$

For the basic exponential function graph, the horizontal axis is the horizontal asymptote. Here, this has been moved down $7$ .

The “inside”, representing the domain, is $x+5$ . This equals $0$ , when $x=-5$ . The exponent is positive for $x>-5$ , since the base is $2 > 1$ , this is the direction of unbounded growth. Therefore, the other direction (left) is where the horizontal asymptote is in effect. Since the coefficient, $\frac {1}{3} > 0$ , the unbounded growth is positive.

At $x=-5$ , we have our one anchor point for the graph. The point is $\left (-5, \frac {1}{3} - 7 \right )$ , which is $\frac {1}{3}$ above the horizontal asymptote, $y = -7$ .

Graph of $y = f(x)$ .

Our graph agrees with our analysis.

The natural or implied domain of $f$ is $\mathbb {R}$ .
$f$ is always increasing.
$f$ has no maximums or minimums.
$\lim \limits _{x \to -\infty } f(x) = -7$
$\lim \limits _{x \to \infty } f(x) = \infty$

Exponential Function

Analyze $B(t) = -2 \, \left ( \frac {2}{3} \right )^{3-t} + 4$

First, observe that $\frac {2}{3} < \answer {1}$ .

Our base is less than $1$ . Therefore, as its exponent gets large and positive, we multiply by more $\frac {2}{3}$ ’s and the overall values get smaller.

Except, the variable, $t$ , in the exponent is multiplied by $-1$ . Therefore, we need $t$ to get large and negative in order for the exponent to get large and positve.

$\left ( \frac {2}{3} \right )^{3-t}$ decays when $t$ becomes more negative.
$\left ( \frac {2}{3} \right )^{3-t}$ grows when $t$ becomes more positive.

The exponential stem of $B(t)$ is $\left ( \frac {2}{3} \right )^{-t}$ , which is a transformed version of the basic exponential function model $M(t) = \left ( \frac {2}{3} \right )^{t}$ .

When $t < 0$ , then $-t > 0$ and we get $\left ( \frac {2}{3} \right )^{-t} = \left ( \frac {2}{3} \right )^{positive}$ and the stem is becoming smaller, approaching $0$ .

$\lim \limits _{t \to -\infty } \left ( \frac {2}{3} \right )^{-t} = 0$

When $t > 0$ , then $-t < 0$ and we get $\left ( \frac {2}{3} \right )^{-t} = \left ( \frac {2}{3} \right )^{negative}$ and the stem is becoming larger.

$\lim \limits _{t \to \infty } \left ( \frac {2}{3} \right )^{-t} = \infty$

Next, we have a negative leading coefficient.

Since $-2 < 0$ , the values of $-2 \, \left ( \frac {2}{3} \right )^{power}$ are always negative.

Finally, we also have two shifts:

$\blacktriangleright$ Vertical Shift

Adding $4$ to the outside shifts the graph vertically up $4$ . The asymptote is $y = 4$ and

$\lim \limits _{t \to -\infty } B(t) = 4$

$\blacktriangleright$ Horizontal Shift

Our exponent is $3 - t = -t + 3$ . Our anchor point for graphing is associated with the exponent equalling $0$ .

$3-t=0$ when $t=3$ . Our one anchor point is shifted over to $3$ . Multipying by $-2$ , means the dot is $2$ away from the horizontal asymptote, which is now $y=4$ .

Graph of $y = B(t)$ .

Our graphical analysis tells us that

The natural or implied domain of $B$ is $\mathbb {R}$ .
$B$ is always decreasing.
$B$ has no maximums or minimums.
$\lim \limits _{t \to -\infty } B(t) = 4$
$\lim \limits _{t \to \infty } B(t) = -\infty$

Exponential Function

Analyze $K(f) = 3^{5-f} - 5$

The exponent gets big and positive when $f$ gets big and positivenegative .

The graph will grow to the rightleft .

The graph will approach the asymptote to the rightleft .

Our one anchor point moves to $\left (\answer {5}, \answer {-4}\right )$ .

The graph will become unbounded updown .

Graph of $y = K(f)$ .

Our graphical analysis tells us that

The natural or implied domain of $K$ is $\mathbb {R}$ .
$K$ is always decreasing.
$K$ has no maximums or minimums.
$\lim \limits _{f \to -\infty } K(f) = \infty$
$\lim \limits _{f \to \infty } K(f) = -5$

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