# 0.8 Exponential functions and graphs  (Page 2/2)

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Therefore, if $a<0$ , then the range is $\left(-\infty ,q\right)$ , meaning that $f\left(x\right)$ can be any real number less than $q$ . Equivalently, one could write that the range is $\left\{y\in \mathbb{R}:y .

For example, the domain of $g\left(x\right)=3·{2}^{x+1}+2$ is $\left\{x:x\in \mathbb{R}\right\}$ . For the range,

$\begin{array}{ccc}\hfill {2}^{x+1}& >& 0\hfill \\ \hfill 3·{2}^{x+1}& >& 0\hfill \\ \hfill 3·{2}^{x+1}+2& >& 2\hfill \end{array}$

Therefore the range is $\left\{g\left(x\right):g\left(x\right)\in \left[2,\infty \right)\right\}$ .

## Domain and range

1. Give the domain of $y={3}^{x}$ .
2. What is the domain and range of $f\left(x\right)={2}^{x}$ ?
3. Determine the domain and range of $y={\left(1,5\right)}^{x+3}$ .

## Intercepts

For functions of the form, $y=a{b}^{\left(x+p\right)}+q$ , the intercepts with the $x$ - and $y$ -axis are calculated by setting $x=0$ for the $y$ -intercept and by setting $y=0$ for the $x$ -intercept.

The $y$ -intercept is calculated as follows:

$\begin{array}{ccc}\hfill y& =& a{b}^{\left(x+p\right)}+q\hfill \\ \hfill {y}_{int}& =& a{b}^{\left(0+p\right)}+q\hfill \\ & =& a{b}^{p}+q\hfill \end{array}$

For example, the $y$ -intercept of $g\left(x\right)=3·{2}^{x+1}+2$ is given by setting $x=0$ to get:

$\begin{array}{ccc}\hfill y& =& 3·{2}^{x+1}+2\hfill \\ \hfill {y}_{int}& =& 3·{2}^{0+1}+2\hfill \\ & =& 3·{2}^{1}+2\hfill \\ & =& 3·2+2\hfill \\ & =& 8\hfill \end{array}$

The $x$ -intercepts are calculated by setting $y=0$ as follows:

$\begin{array}{ccc}\hfill y& =& a{b}^{\left(x+p\right)}+q\hfill \\ \hfill 0& =& a{b}^{\left({x}_{int}+p\right)}+q\hfill \\ \hfill a{b}^{\left({x}_{int}+p\right)}& =& -q\hfill \\ \hfill {b}^{\left({x}_{int}+p\right)}& =& -\frac{q}{a}\hfill \end{array}$

Since $b>0$ (this is a requirement in the original definition) and a positive number raised to any power is always positive, the last equation above only has a real solution if either $a<0$ or $q<0$ (but not both). Additionally, $a$ must not be zero for the division to be valid. If these conditions are not satisfied, the graph of the function of the form $y=a{b}^{\left(x+p\right)}+q$ does not have any $x$ -intercepts.

For example, the $x$ -intercept of $g\left(x\right)=3·{2}^{x+1}+2$ is given by setting $y=0$ to get:

$\begin{array}{ccc}\hfill y& =& 3·{2}^{x+1}+2\hfill \\ \hfill 0& =& 3·{2}^{{x}_{int}+1}+2\hfill \\ \hfill -2& =& 3·{2}^{{x}_{int}+1}\hfill \\ \hfill {2}^{{x}_{int}+1}& =& \frac{-2}{2}\hfill \end{array}$

which has no real solution. Therefore, the graph of $g\left(x\right)=3·{2}^{x+1}+2$ does not have a $x$ -intercept. You will notice that calculating $g\left(x\right)$ for any value of $x$ will always give a positive number, meaning that $y$ will never be zero and so the graph will never intersect the $x$ -axis.

## Intercepts

1. Give the y-intercept of the graph of $y={b}^{x}+2$ .
2. Give the x- and y-intercepts of the graph of $y=\frac{1}{2}{\left(1,5\right)}^{x+3}-0,75$ .

## Asymptotes

Functions of the form $y=a{b}^{\left(x+p\right)}+q$ always have exactly one horizontal asymptote.

When examining the range of these functions, we saw that we always have either $y or $y>q$ for all input values of $x$ . Therefore the line $y=q$ is an asymptote.

For example, we saw earlier that the range of $g\left(x\right)=3·{2}^{x+1}+2$ is $\left(2,\infty \right)$ because $g\left(x\right)$ is always greater than 2. However, the value of $g\left(x\right)$ can get extremely close to 2, even though it never reaches it. For example, if you calculate $g\left(-20\right)$ , the value is approximately 2.000006. Using larger negative values of $x$ will make $g\left(x\right)$ even closer to 2: the value of $g\left(-100\right)$ is so close to 2 that the calculator is not precise enough to know the difference, and will (incorrectly) show you that it is equal to exactly 2.

From this we deduce that the line $y=2$ is an asymptote.

## Asymptotes

1. Give the equation of the asymptote of the graph of $y={3}^{x}-2$ .
2. What is the equation of the horizontal asymptote of the graph of $y=3{\left(0,8\right)}^{x-1}-3$ ?

## Sketching graphs of the form $f\left(x\right)=a{b}^{\left(x+p\right)}+q$

In order to sketch graphs of functions of the form, $f\left(x\right)=a{b}^{\left(x+p\right)}+q$ , we need to determine four characteristics:

1. domain and range
2. $y$ -intercept
3. $x$ -intercept

For example, sketch the graph of $g\left(x\right)=3·{2}^{x+1}+2$ . Mark the intercepts.

We have determined the domain to be $\left\{x:x\in \mathbb{R}\right\}$ and the range to be $\left\{g\left(x\right):g\left(x\right)\in \left(2,\infty \right)\right\}$ .

The $y$ -intercept is ${y}_{int}=8$ and there is no $x$ -intercept. Graph of g ( x ) = 3 · 2 x + 1 + 2 .

## Sketching graphs

1. Draw the graphs of the following on the same set of axes. Label the horizontal asymptotes and y-intercepts clearly.
1. $y={b}^{x}+2$
2. $y={b}^{x+2}$
3. $y=2{b}^{x}$
4. $y=2{b}^{x+2}+2$
1. Draw the graph of $f\left(x\right)={3}^{x}$ .
2. Explain where a solution of ${3}^{x}=5$ can be read off the graph.

## End of chapter exercises

1. The following table of values has columns giving the $y$ -values for the graph $y={a}^{x}$ , $y={a}^{x+1}$ and $y={a}^{x}+1$ . Match a graph to a column.
 $x$ A B C -2 7,25 6,25 2,5 -1 3,5 2,5 1 0 2 1 0,4 1 1,4 0,4 0,16 2 1,16 0,16 0,064
2. The graph of $f\left(x\right)=1+a.{2}^{x}$ (a is a constant) passes through the origin.
1. Determine the value of $a$ .
2. Determine the value of $f\left(-15\right)$ correct to FIVE decimal places.
3. Determine the value of $x$ , if $P\left(x;0,5\right)$ lies on the graph of $f$ .
4. If the graph of $f$ is shifted 2 units to the right to give the function $h$ , write down the equation of $h$ .
3. The graph of $f\left(x\right)=a.{b}^{x}\phantom{\rule{3.33333pt}{0ex}}\left(a\ne 0\right)$ has the point P(2;144) on $f$ .
1. If $b=0,75$ , calculate the value of $a$ .
2. Hence write down the equation of $f$ .
3. Determine, correct to TWO decimal places, the value of $f\left(13\right)$ .
4. Describe the transformation of the curve of $f$ to $h$ if $h\left(x\right)=f\left(-x\right)$ .

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