Chapter 9.2, Problem 19PPS

To determine

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

Answer to Problem 19PPS

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

Explanation of Solution

Given:

The given equation is $x=\frac{2}{3}{y}^2-4x+12$ .

Calculation:

Consider the given equation is,

  $x=\frac{2}{3}{y}^2-4x+12$

Write the above equation in the vertex form as,

  $\begin{array}{c}x=\frac{2}{3}{y}^2-4x+12\\ =\frac{2}{3}\left(y^2-6y+9\right)+12-\frac{2}{3}\left(9\right)\\ =\frac{2}{3}{\left(y-3\right)}^2+6\end{array}$

Consider the general vertex form is,

  $x=a{\left(x-k\right)}^2+h$

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

Then, the vertex form is $\left(h,k\right)=\left(6,3\right)$

The axis of symmetry is $y=3$ .

The value of $a=\frac{2}{3}$ . As it is positive the parabola open upwards.

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