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Use the regression feature to find an exponential function that best fits the data in the table.
$f(x)=731.92{(0.738)}^{x}$
Write the exponential function as an exponential equation with base $\text{\hspace{0.17em}}e.$
Use the intersect feature to find the value of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for which $\text{\hspace{0.17em}}f(x)=250.$
For the following exercises, refer to [link] .
x | f(x) |
1 | 5.1 |
2 | 6.3 |
3 | 7.3 |
4 | 7.7 |
5 | 8.1 |
6 | 8.6 |
Use a graphing calculator to create a scatter diagram of the data.
Use the LOGarithm option of the REGression feature to find a logarithmic function of the form $\text{\hspace{0.17em}}y=a+b\mathrm{ln}\left(x\right)\text{\hspace{0.17em}}$ that best fits the data in the table.
Use the logarithmic function to find the value of the function when $\text{\hspace{0.17em}}x=10.$
$f(10)\approx 9.5$
Graph the logarithmic equation on the scatter diagram.
Use the intersect feature to find the value of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for which $\text{\hspace{0.17em}}f(x)=7.$
When $\text{\hspace{0.17em}}f(x)=7,$ $x\approx \mathrm{2.7.}$
For the following exercises, refer to [link] .
x | f(x) |
1 | 7.5 |
2 | 6 |
3 | 5.2 |
4 | 4.3 |
5 | 3.9 |
6 | 3.4 |
7 | 3.1 |
8 | 2.9 |
Use a graphing calculator to create a scatter diagram of the data.
Use the LOGarithm option of the REGression feature to find a logarithmic function of the form $\text{\hspace{0.17em}}y=a+b\mathrm{ln}\left(x\right)\text{\hspace{0.17em}}$ that best fits the data in the table.
$f(x)=7.544-2.268\mathrm{ln}(x)$
Use the logarithmic function to find the value of the function when $\text{\hspace{0.17em}}x=10.$
Use the intersect feature to find the value of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for which $\text{\hspace{0.17em}}f(x)=8.$
For the following exercises, refer to [link] .
x | f(x) |
1 | 8.7 |
2 | 12.3 |
3 | 15.4 |
4 | 18.5 |
5 | 20.7 |
6 | 22.5 |
7 | 23.3 |
8 | 24 |
9 | 24.6 |
10 | 24.8 |
Use a graphing calculator to create a scatter diagram of the data.
Use the LOGISTIC regression option to find a logistic growth model of the form $\text{\hspace{0.17em}}y=\frac{c}{1+a{e}^{-bx}}\text{\hspace{0.17em}}$ that best fits the data in the table.
To the nearest whole number, what is the predicted carrying capacity of the model?
Use the intersect feature to find the value of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for which the model reaches half its carrying capacity.
When $\text{\hspace{0.17em}}f(x)=12.5,$ $x\approx \mathrm{2.1.}$
For the following exercises, refer to [link] .
$x$ | $f\left(x\right)$ |
0 | 12 |
2 | 28.6 |
4 | 52.8 |
5 | 70.3 |
7 | 99.9 |
8 | 112.5 |
10 | 125.8 |
11 | 127.9 |
15 | 135.1 |
17 | 135.9 |
Use a graphing calculator to create a scatter diagram of the data.
Use the LOGISTIC regression option to find a logistic growth model of the form $\text{\hspace{0.17em}}y=\frac{c}{1+a{e}^{-bx}}\text{\hspace{0.17em}}$ that best fits the data in the table.
$f(x)=\frac{136.068}{1+10.324{e}^{-0.480x}}$
Graph the logistic equation on the scatter diagram.
To the nearest whole number, what is the predicted carrying capacity of the model?
about $136$
Use the intersect feature to find the value of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for which the model reaches half its carrying capacity.
Recall that the general form of a logistic equation for a population is given by $\text{\hspace{0.17em}}P(t)=\frac{c}{1+a{e}^{-bt}},$ such that the initial population at time $\text{\hspace{0.17em}}t=0\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}P(0)={P}_{0}.\text{\hspace{0.17em}}$ Show algebraically that $\text{\hspace{0.17em}}\frac{c-P(t)}{P(t)}=\frac{c-{P}_{0}}{{P}_{0}}{e}^{-bt}.$
Working with the left side of the equation, we see that it can be rewritten as $\text{\hspace{0.17em}}a{e}^{-bt}:$
$\frac{c-P(t)}{P(t)}=\frac{c-\frac{c}{1+a{e}^{-bt}}}{\frac{c}{1+a{e}^{-bt}}}=\frac{\frac{c\left(1+a{e}^{-bt}\right)-c}{1+a{e}^{-bt}}}{\frac{c}{1+a{e}^{-bt}}}=\frac{\frac{c\left(1+a{e}^{-bt}-1\right)}{1+a{e}^{-bt}}}{\frac{c}{1+a{e}^{-bt}}}=1+a{e}^{-bt}-1=a{e}^{-bt}$
Working with the right side of the equation we show that it can also be rewritten as $\text{\hspace{0.17em}}a{e}^{-bt}.\text{\hspace{0.17em}}$ But first note that when $\text{\hspace{0.17em}}t=0,$ $\text{\hspace{0.17em}}{P}_{0}=\frac{c}{1+a{e}^{-b(0)}}=\frac{c}{1+a}.\text{\hspace{0.17em}}$ Therefore,
$\frac{c-{P}_{0}}{{P}_{0}}{e}^{-bt}=\frac{c-\frac{c}{1+a}}{\frac{c}{1+a}}{e}^{-bt}=\frac{\frac{c\left(1+a\right)-c}{1+a}}{\frac{c}{1+a}}{e}^{-bt}=\frac{\frac{c\left(1+a-1\right)}{1+a}}{\frac{c}{1+a}}{e}^{-bt}=\left(1+a-1\right){e}^{-bt}=a{e}^{-bt}$
Thus, $\text{\hspace{0.17em}}\frac{c-P(t)}{P(t)}=\frac{c-{P}_{0}}{{P}_{0}}{e}^{-bt}.$
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