Chapter 2.1, Problem 39E

To determine

To find: The standard form of the quadratic function and the graph for it, also identify the graph, vertex axis of symmetry and x intercepts.

Answer to Problem 39E

The quadratic function in the standard form is $f\left(x\right)=2{\left(x-4\right)}^2$ , the vertex of the graph is $\left(4,0\right)$ and its axis of symmetry is $x=4$ , the $x$ intercepts are $\left(4,0\right)$ and the graph is shown in Figure 1

Explanation of Solution

Given:

The given function is $f\left(x\right)=2x^2-16x+32$ .

Calculation:

Consider the quadratic function is,

  $f\left(x\right)=a{\left(x-h\right)}^2+k$

The vertex of the above parabola for the standard form $f\left(x\right)=a{\left(x-h\right)}^2+k$ is,

  $\left(h,k\right)$

Consider the given equation is,

  $\begin{array}{c}f\left(x\right)=2x^2-16x+32\\ =2\left(x^2-8x+16\right)\\ =2{\left(x-4\right)}^2\end{array}$

As the leading coefficient $a>0$ therefore, the parabola will open in the upward direction.

The required sketch for the graph is shown in Figure 1

  

Figure 1

From the above equation the vertex of the function $f\left(x\right)=2{\left(x-4\right)}^2$ is,

  $\left(h,k\right)=\left(4,0\right)$

Thus, the axis of symmetry of the graph is $x=4$ .

To find the $x$ intercept of the function $f\left(x\right)=2x^2-16x+32$ equate the expression equal to zero as,

  $\begin{array}{l}2x^2-16x+32=0\\ 2\left(x^2-8x+16\right)=0\\ 2{\left(x-4\right)}^2=0\\ x=4\end{array}$

The $x$ intercept of the equation is at $\left(4,0\right)$ .