Chapter 9.2, Problem 16PPS

To determine

To find: The vertex form of the quadratic equation, identify the vertex, axis of symmetry and direction of opening.

Answer to Problem 16PPS

The vertex form of the equation is $y=-6{\left(x+3\right)}^2+46$ , the vertex is $\left(h,k\right)=\left(-3,46\right)$ and the axis of symmetry is $x=-3$ . As $a=-6$ is negative so parabola open downwards.

Explanation of Solution

Given:

The given equation is $y=-6x^2-36x-8$ .

Calculation:

Consider the given equation is,

  $y=-6x^2-36x-8$

Write the above equation in the vertex form as,

  $\begin{array}{c}y=-6x^2-36x-8\\ =-6\left(x^2+6x+9\right)+54-8\\ =-6{\left(x+3\right)}^2+46\end{array}$

Consider the general vertex form is,

  $y=a{\left(x-h\right)}^2+k$

Here, $\left(h,k\right)$ are the vertex, $x$ is the axis of symmetry.

Then, the vertex form is $\left(h,k\right)=\left(-3,46\right)$

The axis of symmetry is $x=-3$ .

The value of $a=1$ . As it is positive the parabola open upwards.

Thus, vertex form of the equation is $y=-6{\left(x+3\right)}^2+46$ , the vertex is $\left(h,k\right)=\left(-3,46\right)$ and the axis of symmetry is $x=-3$ . As $a=-6$ is negative so parabola open downwards.