Practice 2 Solutions

5. $f\left(x\right)={\left(3\right)}^x, g\left(x\right)=2{\left(3\right)}^{x–4}+6$

g(x) is vertically stretched and translated right 4 units and up 6 units.

6. $f\left(x\right)={\left(\frac{4}{5}\right)}^{–x}, g\left(x\right)={\left(\frac{5}{4}\right)}^x$

f(x) and g(x) are the same graph.

7. $f\left(x\right)={\left(7\right)}^x–3, g\left(x\right)=–4{\left(7\right)}^{–x}+1$

f(x) is a growth function, and g(x) is a reflected decay function. g(x) is translated up 4 units from f(x).

8. $f\left(x\right)={\left(6\right)}^{x–5}, g\left(x\right)={\left(6\right)}^{x+5}$

g(x) is translated left 10 units from f(x).

9. $f\left(x\right)=–{\left(2\right)}^x+4$

$\mathrm{As} x\to +\infty , f\left(x\right)\to –\infty , \mathrm{and} \mathrm{as} x\to –\infty , f\left(x\right)\to 4$

10. $g\left(x\right)=e^{\left(x+1\right)}–3$

$a=1, b=e, h=–1, k=–3$

Sample table:

$\mathrm{As} x\to +\infty , g\left(x\right)\to +\infty , \mathrm{and} \mathrm{as} x\to –\infty , g\left(x\right)\to –3$

11. $h\left(x\right)={\left(\frac{1}{4}\right)}^x$

$a=1, b=\frac{1}{4}, h=0, k=0$

Sample table:

$\mathrm{As} x\to +\infty , h\left(x\right)\to 0, \mathrm{and} \mathrm{as} x\to –\infty , h\left(x\right)\to +\infty$

12. $p\left(x\right)={\left(\frac{5}{4}\right)}^{–x}$

$a=1, b=\frac{4}{5}, h=0, k=0$

Sample table:

$\mathrm{As} x\to +\infty , p\left(x\right)\to 0, \mathrm{and} \mathrm{as} x\to –\infty , p\left(x\right)\to +\infty$

13. $f\left(x\right)=–13{\left(\frac{1}{4}\right)}^{–x}–6$

$$\begin{gathered} {\left(\frac{1}{4}\right)}^{–x}=4^x \\ a=–13, b=4, h=0, k=–6 \end{gathered}$$

This function is a growth function. It is reflected over the x-axis because a is negative, with a vertical stretch. It is translated 6 units down.

$$\begin{gathered} \mathrm{Domain}: \left\{x|x\in \mathrm{ℝ}\right\} \\ \mathrm{Range}: \left\{y|y<–6\right\} \end{gathered}$$

14. $g\left(x\right)=15{\left(e\right)}^x$

$a=15, b=e, h=0, k=0$

This function is a growth function with a vertical stretch.

$$\begin{gathered} \mathrm{Domain}: \left\{x|x\in \mathrm{ℝ}\right\} \\ \mathrm{Range}: \left\{y|y>0\right\} \end{gathered}$$

15. $h\left(x\right)=9{\left(\frac{1}{8}\right)}^{x–1}+2$

$a=9, b=\frac{1}{8}, h=1, k=2$

This function is a decay function with a vertical stretch that is translated one unit to the right and two units up.

$$\begin{gathered} \mathrm{Domain}: \left\{x|x\in \mathrm{ℝ}\right\} \\ \mathrm{Range}: \left\{y|y>2\right\} \end{gathered}$$

16. $q\left(x\right)={\left(1\right)}^x$

b = 1

This function is neither a decay nor a growth function because $b=1.$ One raised to any power is equal to one.

$$\begin{gathered} \mathrm{Domain}: \left\{x|x\in \mathrm{ℝ}\right\} \\ \mathrm{Range}: \left\{y|y=1\right\} \end{gathered}$$

17. (–2, 1) and (1, 8)

$\begin{array}{rcl}y& =& a{b}^x\\ 1& =& a{b}^{–2}\\ b^2& =& a\\ 8& =& b^2{\left(b\right)}^1\\ 8& =& b^3\\ b& =& 2\\ a& =& {\left(2\right)}^2=4\end{array}$

$y=4{\left(2\right)}^x$

18. (1, –3) and (–2, –81)

$\begin{array}{rcl}y& =& a{b}^x\\ –3& =& a{b}^1\\ a& =& –\frac{3}{b}\\ –81& =& \left(–\frac{3}{b}\right)\left(b^{–2}\right)\\ 81& =& \frac{3}{b^3}\\ 81b^3& =& 3\\ b^3& =& \frac{1}{27}\\ b& =& \frac{1}{3}\\ a& =& –\frac{3}{{\displaystyle \frac{1}{3}}}\\ a& =& –9\end{array}$

$y=–9{\left(\frac{1}{3}\right)}^x$

19. $\left(2,\frac{ 2}{3}\right)$ and $\left(–1, 144\right)$

$\begin{array}{rcl}y& =& a{b}^x\\ \frac{2}{3}& =& a{b}^2\\ a& =& \frac{2}{3b^2}\\ 144& =& \left(\frac{2}{3b^2}\right)b^{–1}\\ 144& =& \frac{2}{3b^3}\\ 144b^3& =& \frac{2}{3}\\ b^3& =& \frac{1}{216}\\ b& =& \frac{1}{6}\\ a& =& \frac{2}{3{\left({\displaystyle \frac{1}{6}}\right)}^2}=\frac{2}{{\displaystyle \frac{1}{12}}}=24\end{array}$

$y=24{\left(\frac{1}{6}\right)}^x$

20. One year ago, Ringo purchased a $0.20 comic book at a yard sale. When he showed the comic to a collector, he was told to hold onto it because it would increase in value. In two more years, the comic is worth $25. Write an exponential equation to model the value of the comic book.

$\begin{array}{rcl}& & \left(–1, 0.20\right), \left(2, 25\right)\\ y& =& a{b}^x\\ 25& =& a{b}^2\\ a& =& \frac{25}{b^2}\\ 0.2& =& \left(\frac{25}{b^2}\right)b^{–1}\\ 0.2& =& \frac{25}{b^3}\\ 0.2b^3& =& 25\\ b^3& =& 125\\ b& =& 5\\ a& =& \frac{25}{5^2}=1\\ & & \end{array}$

$y=5^x$