Chapter 8.4, Problem 53E

To determine

To graph function $g(x)=f(x-4),\text{where}f(x)={\left(x+3\right)}^2+5$

Explanation of Solution

Given information :

Given functions $g(x)=f(x-4),\text{where}f(x)={\left(x+3\right)}^2+5$

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 $f(x)={\left(x+3\right)}^2+5$ with $f(x)=m{(x-h)}^2+k$ , we get $h=-3,k=5$

Thus, the vertex of $f(x)$ is $(-3,5)$ and the axis of symmetry $x=-3$

Given function $g(x)=f(x-4)$ is in the form of $y=f(x-h)$ , where $h=4$ .

So, the graph of $g(x)$ is a horizontal translation of $4$ units right of the graph of $f(x)$ ,

To graph $g(x)$ , add $4$ from the $x-$ coordinates of the points on the graph of $f(x)$ .

Thus, the vertex and axis of symmetry of $g(x)$ are $(1,5)$ and $x=1$ respectively.

Graph: The graph of $g(x)=f(x-4)$ is

  

Interpretation: From above graph, we observe the vertex and axis of symmetry of $g(x)$ are $(1,5)$ and $x=1$ respectively .