Chapter 9.2, Problem 17PPS

To determine

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

Answer to Problem 17PPS

The vertex form of the equation is $y=-3{\left(x+1.5\right)}^2+0.75$ , the vertex is $\left(h,k\right)=\left(-1.5,0.75\right)$ and the axis of symmetry is $x=0.75$ . As $a=-3$ is negative so parabola open downwards.

Explanation of Solution

Given:

The given equation is $y=-3x^2-9x-6$ .

Calculation:

Consider the given equation is,

  $y=-3x^2-9x-6$

Write the above equation in the vertex form as,

  $\begin{array}{c}y=-3x^2-9x-6\\ =-3\left(x^2+x+{1.5}^2\right)-6-3{\left(1.5\right)}^2\\ =-3{\left(x+1.5\right)}^2+0.75\end{array}$

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 $\left(h,k\right)=\left(-1.5,0.75\right)$

The axis of symmetry is $x=0.75$ .

The value of $a=-3$ . As it is positive the parabola open upwards.

Thus, vertex form of the equation is $y=-3{\left(x+1.5\right)}^2+0.75$ , the vertex is $\left(h,k\right)=\left(-1.5,0.75\right)$ and the axis of symmetry is $x=0.75$ . As $a=-3$ is negative so parabola open downwards.