Chapter 9.2, Problem 18PPS

To determine

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

Answer to Problem 18PPS

The vertex form of the equation is $x=\frac{1}{3}{\left(y-4.5\right)}^2-2.75$ , the vertex is $\left(h,k\right)=\left(-2.75,4.5\right)$ and the axis of symmetry is $y=4.5$ . As $a=\frac{1}{3}$ is positive so parabola open upwards.

Explanation of Solution

Given:

The given equation is $x=\frac{1}{3}{y}^2-3x+4$ .

Calculation:

Consider the given equation is,

  $x=\frac{1}{3}{y}^2-3x+4$

Write the above equation in the vertex form as,

  $\begin{array}{c}x=\frac{1}{3}\left(y^2-9x\right)+4\\ =\frac{1}{3}\left(y^2-9x+{4.5}^2\right)+4-\frac{1}{3}{\left(4.5\right)}^2\\ =\frac{1}{3}{\left(y-4.5\right)}^2-2.75\end{array}$

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 $\left(h,k\right)=\left(-2.75,4.5\right)$

The axis of symmetry is $y=4.5$ .

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

Thus, vertex form of the equation is $x=\frac{1}{3}{\left(y-4.5\right)}^2-2.75$ , the vertex is $\left(h,k\right)=\left(-2.75,4.5\right)$ and the axis of symmetry is $y=4.5$ . As $a=\frac{1}{3}$ is positive so parabola open upwards.