Chapter 9.2, Problem 4CYU

To determine

To write: The equation in standard form and identify the vertex, axis of symmetry and direction of opening of the parabola.

Answer to Problem 4CYU

The standard equation is $x=-3{\left(y+2\right)}^2+30$ , vertex is $\left(30,-2\right)$ , axis of symmetry is $y=2$ and parabola opens towards left.

Explanation of Solution

Given:

The given equation is $x+3y^2+12y=18$ .

Calculation:

Rewrite the given quadratic equation in the standard form by completing the square.

  $\begin{array}{c}x+3y^2+12y=18\\ x=-3y^2-12y+18\\ x=-3\left(y^2+4y+4\right)+12+18\\ x=-3{\left(y+2\right)}^2+30\end{array}$

Therefore, the standard form of the given quadratic equation is $x=-3{\left(y+2\right)}^2+30$ .

Compare the equation $x={\left(y-4\right)}^2-27$ with the standard equation $x=a{\left(y-k\right)}^2+h$ .

  $\begin{array}{c}a=-3\\ h=30\\ k=-2\end{array}$

Therefore, the vertex of the given equation is $\left(30,-2\right)$ .

Axis of symmetry will be given shown below.

  $\begin{array}{c}y=k\\ =2\end{array}$

Therefore, the axis of symmetry is $y=2$ .

As it is found that the value of $a=-3$ is positive so by definition the parabola opens towards left.

Therefore, the standard equation is $x=-3{\left(y+2\right)}^2+30$ , vertex is $\left(30,-2\right)$ , axis of symmetry is $y=2$ and parabola opens towards left.