Chapter 9.3, Problem 13PPS

To determine

Describe how the graph of each function is related to the graph of $f(x)=x^2\text{and}h(x)=3+\frac{5}{2}{x}^2$

Answer to Problem 13PPS

The graph of $h(x)=3+\frac{5}{2}{x}^2$ is the graph of $f(x)=x^2$ vertically stretched by a factor $\frac{5}{2}$ and shifted up 3 units.

Explanation of Solution

Given: The functions $f(x)=x^2\text{and}h(x)=3+\frac{5}{2}{x}^2$

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 $a=\frac{5}{2}\text{and}c=+3$. When c is positive, it translated up and if c is negative, it translated down. Since $c=+3$,so the graph is translated 3 units up. Since $\left|a\right|>1$ and c is positive, the graph of $h(x)=3+\frac{5}{2}{x}^2$ is the graph of $f(x)=x^2$ vertically stretched by a factor $\frac{5}{2}$ and shifted up 3 units.

Thus, the graph of $h(x)=3+\frac{5}{2}{x}^2$ is the graph of $f(x)=x^2$ vertically stretched by a factor $\frac{5}{2}$ and shifted up 3 units.