Chapter 1.6, Problem 8QR

To determine

To show: Thefunction $f\left(x\right)=2x^2-3$ is an even function. Also, explain the reason that graph is symmetric about the y -axis.

Explanation of Solution

Given information:The given function is $f\left(x\right)=2x^2-3$ .

Calculation:

A function $f$ is said to be an even function if $f\left(-x\right)=f\left(x\right)$ for all $x$ in its domain.

Substitute $-x$ for $x$ in the function $f\left(x\right)=2x^2-3$ .

  $\begin{array}{c}f\left(-x\right)=2{\left(-x\right)}^2-3\\ =2x^2-3\\ =f\left(x\right)\end{array}$

So, the given function is an even function.

For any value of $a$ in its domain $f\left(-a\right)=f\left(a\right)$ . So, the image of point $\left(a,f\left(a\right)\right)$ along the y -axis is the point $\left(-a,f\left(a\right)\right)$ . Every point on the graph is paired with its symmetric point about the y -axis it means that the graph is symmetric about the y -axis.

Hence, it is proved that the function $f\left(x\right)=2x^2-3$ is an even function.