Chapter 9.4, Problem 62SR

To determine

Describe how the graph of each function is related to the graph of $f(x)=x^2\text{and}h(x)=-1-\frac{3}{2}{x}^2$

Answer to Problem 62SR

The graph of $h(x)=-1-\frac{3}{2}{x}^2$ is the graph of $f(x)=x^2$ reflected across the x − axis, vertically stretched by a factor of $\frac{3}{2}$ and translated down 1 units.

Explanation of Solution

Given: The functions $f(x)=x^2\text{and}h(x)=-1-\frac{3}{2}{x}^2$

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 $f(x)=a{x}^2+c$,where $a=-\frac{3}{2}\text{and}c=-1$ Since $a<0$,negative, the graph is reflected across the x − axis and $\left|a\right|>1$ the graph is vertically stretched by a factor $\frac{3}{2}$. Since $c=-1$,the graph is translated down 1 unit. So, the graph of $h(x)=-1-\frac{3}{2}{x}^2$ is the graph of $f(x)=x^2$ reflected across the x − axis, vertically stretched by a factor of $\frac{3}{2}$ and translated down 1 units.

Thus, The graph of $h(x)=-1-\frac{3}{2}{x}^2$ is the graph of $f(x)=x^2$ reflected across the x − axis, vertically stretched by a factor of $\frac{3}{2}$ and translated down 1 units.