Practice 1 Solutions

1. $f\left(x\right)={\left(\frac{3}{7}\right)}^x$

$b=\frac{3}{7}$, Decay

2. $g\left(x\right)=9^x$

b = 9, Growth

4. $v\left(x\right)={\left(0.3\right)}^{–x}$

${\left(0.3\right)}^{–x}={\left(\frac{1}{0.3}\right)}^x={\left(\frac{10}{3}\right)}^x$

$b=\frac{10}{3}$, Growth

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

g(x) is reflected across the x-axis from f(x).

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

f(x) is a decay function, and g(x) is a growth function.

7. $f\left(x\right)= 5^x, g\left(x\right)=5^{x–1}+2$

g(x) is translated one unit to the right and two units up from f(x).

8. $f\left(x\right)=8^{x+2} , g\left(x\right)=2{\left(8\right)}^{x+5}$

g(x) is vertically stretched and translated 3 units left from f(x).

11. $h\left(x\right)={\left(1.5\right)}^{–x}$

$a=1, b=\frac{2}{3}, h=0, k=0$

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. $f\left(x\right)=5^{x+2}–3$

$a=1, b=5, h=–2, k=–3$

Sample table:

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

13. $f\left(x\right)=12{\left(e\right)}^{–x} +4$

$a=12, b=\frac{1}{e}, h=0, k=4$

This function is a decay function because the base is between 0 and 1 when the exponent is positive.

$$\begin{gathered} \mathrm{Domain}: \left\{x|x\in \mathrm{ℝ}\right\} \\ \mathrm{Range}: \left\{y|y>4\right\} \end{gathered}$$

14. $g\left(x\right)= –32{\left(7\right)}^{x+3} –2$

$a=–32, b=7, h=–3, k=–2$

This function is a growth function. It is reflected over the x-axis because a is negative.

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

15. $h\left(x\right)= 4{\left(5\right)}^x$

$a=4, b=5, h=0, k=0$

This function is a growth function.

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

16. $r\left(x\right)=–10{\left(\frac{2}{5}\right)}^{–x}–7$

$$\begin{gathered} {\left(\frac{2}{5}\right)}^{–x}={\left(\frac{5}{2}\right)}^x \\ a=–10, b=\frac{5}{2}, h=0, k=–7 \end{gathered}$$

This function is a growth function. It is reflected over the x-axis because a is negative.

$$\begin{gathered} \mathrm{Domain}: \left\{x|x\in \mathrm{ℝ}\right\} \\ \mathrm{Range}: \left\{y|y<–7\right\} \end{gathered}$$

17. When planted from seed, the sprouting rate of radish seedlings appears exponential. At the end of week zero, two radish seedlings are visible. After two weeks, eighteen seedlings are visible. Write the exponential equation that models the growth of the seedlings. (x: time, y: seedlings)

$\begin{array}{rcl}y& =& a{b}^x\\ 2& =& a{b}^0\\ 2& =& a\\ 18& =& 2b^2\\ 9& =& b^2\\ b& =& 3\\ b& \ne & –3\end{array}$

Because the base of an exponential function cannot be negative

The equation to model the growth of seedlings is$y=2{\left(3\right)}^x.$

18. (0, –5) and $\left(2, –\frac{5}{16}\right)$

$\begin{array}{rcl}y& =& a{b}^x\\ –5& =& a{b}^0\\ a& =& –5\\ –\frac{5}{16}& =& –5b^2\\ –5& =& –80b^2\\ b^2& =& \frac{1}{16}\\ b& =& \frac{1}{4}\end{array}$

$y=–5{\left(\frac{1}{4}\right)}^x$

19. (1, 4) and $\left(–2,\frac{ 1}{16}\right)$

$\begin{array}{rcl}y& =& a{b}^x\\ 4& =& a{b}^1\\ a& =& \frac{4}{b}\\ \frac{1}{16}& =& \left(\frac{4}{b}\right)b^{–2}\\ \frac{1}{16}{b}^3& =& 4\\ b^3& =& 64\\ b& =& 4\\ a& =& \frac{4}{4}=1\end{array}$

$y=4^x$

20. Two hours ago, twelve mold spores were counted under the microscope. One hour ago, six mold spores remained. Write an exponential equation to model the decline in the mold spores. (x: time, y: amount of mold spores)

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

The equation that models the declining mold spores is$y=3{\left(\frac{1}{2}\right)}^x.$