Chapter 2.2, Problem 26E

To determine

To find: To graph the function, label the vertex and axis of symmetry.

Answer to Problem 26E

The vertex of the function $f(x)=6x^2-5$ is $(0,-5)$ .

Axis of symmetry $x=0$ .

  

Explanation of Solution

Given information: $f(x)=6x^2-5$ .

Identifying the vertex:

General form of a quadratic function $y=a{x}^2+bx+c$ ,

From the function $f(x)=6x^2-5$ ,

  $a=6,b=0,c=-5$

The $x-\text{coordinate}$ of a quadratic function is $x=-\frac{b}{2a}$ ,

  $\begin{array}{l}x=-\frac{0}{2\times (6)}\\ x=-\frac{0}{(12)}\\ x=0\end{array}$

Substituting $x=0\text{in}y=6x^2-5$ ,

  $\begin{array}{l}$ y=6{(0)}^2-5$y=0-5\\ y=-5\end{array}$

The vertex of the given quadratic function $(h,k)$ is $(0,-5)$ .

The axis of symmetry is a vertical line which passes through the vertex of the parabola.

The $x-\text{coordinate}$ of the vertex is the equation of axis of symmetry $x=0$ .