Chapter 8.4, Problem 41E

To determine

To graph function $d(x)=4{(x+1)}^2-5$ and compare with the graph of $f(x)=x^2$

Answer to Problem 41E

The graph of $d(x)$ is a horizontal translation of $1$ unit left, a vertical stretch by a factor of $\text{4}\text{units}$ and a vertical translation of $-5\text{units}$ down of the graph of $f(x)=x^2$

Explanation of Solution

Given information :

Given functions $d(x)=4{(x+1)}^2-5,andf(x)=x^2$

As we know that the vertex of the graph of $f(x)=m{(x-h)}^2+k\left(\text{where}m\text{is}\text{constant}\right)$ is $(h,k)$ and the axis of symmetry is $x=h$ .

Now compare given function $f(x)=x^2$ with $f(x)=m{(x-h)}^2+k$ , we get $h=0,k=0$

Thus, the vertex of $f(x)$ is $(0,0)$ and the axis of symmetry $x=0$

Given function $d(x)=4{(x+1)}^2-5,\text{where}\text{f(x)}\text{=}{\text{x}}^2$ is in the form of $y=af(x-h)+k$ , where $h=-1,a=4>1andk=-5$ .

So, the graph of $d(x)$ is a horizontal translation of $1$ unit left, a vertical stretch by a factor of $\text{4}\text{units}$ and a vertical translation of $-5\text{units}$ down of the graph of $f(x)=x^2$ .

Graph: The graph of $d(x)=4{(x+1)}^2-5,andf(x)=x^2$ is

  

Interpretation: From above graph, we observe the graph of $d(x)$ is a horizontal translation of $1$ unit left, a vertical stretch by a factor of $\text{4}\text{units}$ and a vertical translation of $-5\text{units}$ down of the graph of $f(x)=x^2$