Answer
a. See explanations.
b. $40.5\ in$, fits well.
c. $1.0$ inches per month.
d. $0.2$ inches per month. See explanations.
Work Step by Step
a. To obtain the graph of function $f(x)=3.1\sqrt x+19$ from the curve of $y=\sqrt x$, stretch the curve vertically by a factor of 3.1, then shift the result vertically $19\ in$ up.
b. For $x=48$ month, we have $f(x)=3.1\sqrt {48}+19\approx40.5\ in$. Compared to the actual median height of $40.2\ in$, we can see that the model describes the actual height well.
c. We can obtain the average rate of change from $0$ to $10$ month as $R_1=\frac{f(10)-f(0)}{10-0}=\frac{3.1\sqrt {10}}{10}\approx1.0$ inches per month.
d. We can obtain the average rate of change from $50$ to $60$ month as $R_2=\frac{f(60)-f(50)}{60-50}=\frac{3.1(\sqrt {60}-\sqrt {50})}{10}\approx0.2$ inches per month. This growth rate is much smaller than the result from part-c, and this is shown in the graph by the slow increase of the curve when the x-value gets bigger, that is the height growth slows down when the age gets bigger.
