Chapter 9.4, Problem 60SR

To determine

Describe how the graph of each function is related to the graph of $f(x)=x^2\text{and}h(x)=-6+\frac{2}{3}{x}^2$

Answer to Problem 60SR

The graph $h(x)=-6+\frac{2}{3}{x}^2$ is the graph of $y=x^2$ translated 6 units down and compress vertically by a factor $\frac{2}{3}$ .

Explanation of Solution

Given:

The functions $f(x)=x^2\text{and}h(x)=-6+\frac{2}{3}{x}^2$

Concept Used:

  $\begin{array}{l}\text{Vertical}\text{Stretch}\text{and}\text{Vertical}\text{Compression:}\\ \text{y}\text{=}\text{a}\text{f}\text{(x)}\text{,}\text{a}\text{>1}\text{,}\text{Srtetch}\text{the}\text{graph}\text{of}\text{f(x)}\text{vertically}\text{by}\text{a}\text{factor}\text{of}\text{a}\text{.}\\ \text{y}\text{=}\text{a}\text{f}\text{(x)}\text{,}\text{0}\text{<}\text{a}\text{<1}\text{,}\text{Compress}\text{the}\text{graph}\text{of}\text{f(x)}\text{vertically}\text{by}\text{a}\text{factor}\text{of}\text{a}\end{array}$

The function can be written as $f(x)=a{x}^2+c$, the graph stretches or compress the parent graph vertically and translated up or down. If $\left|a\right|<1$,compress the graph f(x) vertically by a factor of $\left|a\right|$ If c is positive, then translated up and if c is negative, then translated down. Since $a=\frac{2}{3}$,the graph of $h(x)=-6+\frac{2}{3}{x}^2$ is the graph of $y=x^2$ vertically compressed by a factor $\frac{2}{3}$. $c=-6$,translated 6 units down. So, the graph $h(x)=-6+\frac{2}{3}{x}^2$ is the graph of $y=x^2$ translated 6 units down and compress vertically by a factor $\frac{2}{3}$.

Thus, the graph $h(x)=-6+\frac{2}{3}{x}^2$ is the graph of $y=x^2$ translated 6 units down and compress vertically by a factor $\frac{2}{3}$ .