Chapter 2.1, Problem 31E

To determine

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

Explanation of Solution

Given data: $y$

The given quadratic function is $f\left(x\right)=-2x^2+16x-31$ .

Concept used:

The standard form of quadratic function $f\left(x\right)=a{\left(x-\text{α}\right)}^2+\text{β}$ represents a parabola havingvertex $\left(\text{α,β}\right)$ . 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 $f\left(x\right)=-2x^2+16x-31$ may be written as

  $f\left(x\right)=-2x^2+16x-31=-2\left(x^2-8x+\frac{31}{2}\right)$

  $f\left(x\right)=-2\left(x^2-8x+16+\frac{31}{2}-16\right)$

  $f\left(x\right)=-2\left(x^2-8x+16\right)-31+32$

  $\Rightarrow f\left(x\right)=-2\cdot {\left(x-4\right)}^2+1$  .........(1)

Now, comparing equation (1) with the function $f(x)=a{\left(x-\text{α}\right)}^{\text{2}}\text{+β}$ , we have $a\text{=-}2$ , $\text{α=}4$ , and $\text{β}=1$ .

Thus, the quadratic function $f\left(x\right)=-2x^2+16x-31$ represents a parabola with vertex $\left(4,1\right)$ .

Since, $a=-2<0$ , hence the parabola has maximum at the vertex so is concave downwards.

The x-intercept of the parabola is given by substituting $f\left(x\right)=0$ and solving this equation for variable “x”.

  $f\left(x\right)=0\Rightarrow -2\cdot {\left(x-4\right)}^2+1=0$

  $\Rightarrow {\left(x-4\right)}^2=\frac{1}{2}$

  $\Rightarrow \left(x-4\right)=\pm \frac{1}{\sqrt{2}}$

  $\Rightarrow x=4-\frac{1}{\sqrt{2}},4+\frac{1}{\sqrt{2}}$

  $\Rightarrow x=3.293,4.707$ .

The curve parabola intersects x-axis at points $\left(3.293,0\right)$ , and $\left(4.707,0\right)$ . Therefore, the intercepts of parabola on x-axis are $OA=3.293$ and $OB=4.707$ .

The graph of the function $f\left(x\right)=-2x^2+16x-31$ (a parabola) is shown in Figure-1 here.

  

We observe that the graph of function $f\left(x\right)=-2x^2+16x-31$ is a downwards concave parabola having vertex $\left(4,1\right)$ and two intercepts $OA=3.293$ and $OB=4.707$ that verify the results described above The graph of the function $f\left(x\right)=-2x^2+16x-31$ is by(i) reflecting the graph of function $h\left(x\right)=x^2$ through x-axis, then (ii) shifting through a distance1 units alongy-axis, and then (ii) shifting it through distance4unitsparallel tox−axisright to the origin.