Chapter 7.3, Problem 44PPS

(a)

To determine

Graph each function on the same set of axes.

Explanation of Solution

Given:

  $\begin{array}{l}f(x)=4\sqrt{x-6}+3\\ g(x)=\sqrt{16x+1}-6\\ h(x)=\sqrt{x+3}+2\end{array}$

Graph:

  

(b)

To determine

Identify the transformation on the graph of the parent function.Find the values caused each transformation.

Explanation of Solution

Given:

  $\begin{array}{l}f(x)=4\sqrt{x-6}+3\\ g(x)=\sqrt{16x+1}-6\\ h(x)=\sqrt{x+3}+2\end{array}$

Calculation:

The parent function is $p(x)=\sqrt{x}$

For $f(x)=4\sqrt{x-6}+3$ , the parent function is shifted 6 units to the right on the x-axis , then stretched vertically by a factor of 4 , then shifted 3 units upward on the y-axis.

For $g(x)=\sqrt{16x+1}-6=4\sqrt{x+\frac{1}{16}}-6$ , the parent function is shifted $\frac{1}{16}$ units to the left on the x-axis , then stretched vertically by a factor of 4, then shifted 6 units downward on the y-axis.

For $h(x)=\sqrt{x+3}+2$ , the parent function is shifted 3 units to the left on the x-axis , then shifted 2 units upward on the y-axis.

(c)

To determine

Find the functions that appear to be stretched or compressed vertically.

Answer to Problem 44PPS

f(x) and g(x) appear stretched vertically by a factor of 4 units.

Explanation of Solution

Given:

  $\begin{array}{l}f(x)=4\sqrt{x-6}+3\\ g(x)=\sqrt{16x+1}-6\\ h(x)=\sqrt{x+3}+2\end{array}$

Calculation: $f(x)=4\sqrt{x-6}+3=\sqrt{16\left(x-6\right)}+3$

  $g(x)=\sqrt{16x+1}-6=\sqrt{16\left(x+\frac{1}{16}\right)}-6$

In both of the functions, since x is multiplied by a factor of 16 , they appear stretched vertically by a factor of $\sqrt{16}=4$ units.

(d)

To determine

The two functions that are stretched appear to be stretched by the same magnitude.Explain how it is possible.

Explanation of Solution

Given:

  $\begin{array}{l}f(x)=4\sqrt{x-6}+3\\ g(x)=\sqrt{16x+1}-6\\ h(x)=\sqrt{x+3}+2\end{array}$

Calculation: In the function f (x), if you take 4 inside the square root , you will get $f(x)=4\sqrt{x-6}+3=\sqrt{16\left(x-6\right)}+3$

Similarly , for g(x) , if you take 16 common , you will get

  $g(x)=\sqrt{16x+1}-6=\sqrt{16\left(x+\frac{1}{16}\right)}-6$

So, in both of the functions, since x is multiplied by a factor of 16 , they appear stretched vertically by a factor of $\sqrt{16}=4$ units.

(e)

To determine

Make a table of the rate of change for all three functions between 8 and 12 as compared to 12 and 16. Explain the generalization about rate of change in square root function that can be made as a result of your findings.

Answer to Problem 44PPS

the rate of change of the square root $a\sqrt{x-b}+c$ functions decreases as the interval of x increases , where a > 0.

Explanation of Solution

Given:

  $\begin{array}{l}f(x)=4\sqrt{x-6}+3\\ g(x)=\sqrt{16x+1}-6\\ h(x)=\sqrt{x+3}+2\end{array}$

Calculation:

Rate of change of a function $y=mx+c$ on interval $[x_1,x_2]$ is given by $\frac{y_2-y_1}{x_2-x_1}$ where $y_1=m{x}_1+c\text{ and }{y}_2=m{x}_2+c$

Function	[8,12]	[12,16]
$f(x)=y=4\sqrt{x-6}+3$	$\begin{array}{l}\frac{4\sqrt{12-6}+3-\left(4\sqrt{8-6}+3\right)}{12-8}\\ =\frac{4\sqrt{6}-4\sqrt{2}}{4}=\sqrt{6}-\sqrt{2}\approx 1.03\end{array}$	$\begin{array}{l}\frac{4\sqrt{16-6}+3-\left(4\sqrt{12-6}+3\right)}{16-12}\\ =\frac{4\sqrt{10}-4\sqrt{6}}{4}=\sqrt{10}-\sqrt{6}\approx 0.71\end{array}$
$g(x)=y=\sqrt{16x+1}-6$	$\begin{array}{l}\frac{\sqrt{16(12)+1}-6-\left(\sqrt{16(8)+1}-6\right)}{12-8}\\ =\frac{\sqrt{193}-\sqrt{129}}{4}\approx 0.633\end{array}$	$\begin{array}{l}\frac{\sqrt{16(16)+1}-6-\left(\sqrt{16(12)+1}-6\right)}{16-12}\\ =\frac{\sqrt{257}-\sqrt{193}}{4}\approx 0.534\end{array}$
$h(x)=y=\sqrt{x+3}+2$	$\begin{array}{l}\frac{\sqrt{12+3}+2-\left(\sqrt{8+3}+2\right)}{12-8}\\ =\frac{\sqrt{15}-\sqrt{11}}{4}\approx 0.13\end{array}$	$\begin{array}{l}\frac{\sqrt{16+3}+2-\left(\sqrt{12+3}+2\right)}{16-12}\\ =\frac{\sqrt{19}-\sqrt{15}}{4}\approx 0.12\end{array}$

So, from the table , we observe that the rate of change of the square root $a\sqrt{x-b}+c$ functions decreases as the interval of x increases , where a > 0.