Chapter 9, Problem 12CR

To determine

The graph of the function.

The axis of symmetry and the vertex.

Answer to Problem 12CR

The vertex of the parabola will be: $\left(-\frac{1}{6},-\frac{61}{12}\right)$

The axis of symmetry will be $x=-\frac{1}{6}$

Explanation of Solution

Given:

The given function is:

  $y=3x^2+x-5$

Calculation:

Compare the equation with $y=a{x}^2+bx+c$

  $a=3,b=1,c=-5$

The x-coordinate of the vertex will be:

  $\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{1}{2\left(3\right)}\\ x=-\frac{1}{6}\end{array}$

The y-coordinate of the vertex will be:

  $\begin{array}{l}y=3x^2+x-5\\ y=3{\left(\frac{1}{6}\right)}^2+\left(\frac{1}{6}\right)-5\\ y=-\frac{61}{12}\end{array}$

The vertex of the parabola will be:

  $\left(-\frac{1}{6},-\frac{61}{12}\right)$

If $V\left(h,k\right)$ is the vertex, then the axis of symmetry will be $x=h$ , therefore, the axis of symmetry will be

  $x=-\frac{1}{6}$

The graph will be: