Chapter 2.1, Problem 24E

To determine

To describe:The graph of given quadraticfunction and identify the vertex, and verify the results with graphing utility.

Explanation of Solution

Given data: $y$

The given quadratic function is $g\left(x\right)=x^2+16x+64$ .

Concept used:

The standard form of quadratic function $g\left(x\right)=a{\left(x-\text{α}\right)}^2+\text{β}$ represents a parabola with vertex is $\left(\text{α,β}\right)$ . If $a>0$ then the parabola has minima at the vertex, so concave upwards, and if $a<0$ then the parabola has maxima at the vertex so concave downwards

Calculations:

The given quadratic function is $g\left(x\right)=x^2+16x+64$ may be written as

  $g\left(x\right)=x^2+16x+64=1\cdot {\left(x+8\right)}^2+0$  .........(1)

Now, comparing equation (1) with the function $g(x)=a{\left(x-\text{α}\right)}^{\text{2}}\text{+β}$ , we have $a\text{=}1$ , $h\text{=-}\text{8}$ , and $k=0$ .

Thus, the quadratic function $g\left(x\right)=x^2+16x+64$ represents a parabola with vertex $\left(-8,0\right)$ .

Since, $a=\text{1>0}$ , hence the parabola is minimum at the vertex so is concave upwards.

The graph of the function $g\left(x\right)=x^2+16x+64$ (a parabola) is shown in Figure-1 here.

We observe that the graph of function $g\left(x\right)=x^2+16x+64$ is an upwards concave parabola with vertex $\left(-8,0\right)$ that verify the results described above.

  

The graph of the function $g\left(x\right)=x^2+16x+64$ is obtained by translatingthe graph of function $g\left(x\right)=x^2$ , through a distance8 unitsalong negative x−axis.