Question

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation

$y^2-3x+6y+12=0$

Short answer

The standard form of the equation is$x=\frac{1}{3}{\left(y+3\right)}^2+1$and represents a parabola.

Step-by-step solution

Step 1. Given Information.

Given to write the equation in standard form and to state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then the equation is to be graphed.

$y^2-3x+6y+12=0$

Step 2. Explanation.

Rewriting the given equation in standard form:

$\begin{array}{l}{y}^2-3x+6y+12=0\\ y^2+6y+9-3x+3=0\\ 3x={\left(y+3\right)}^2+3\\ x=\frac{1}{3}{\left(y+3\right)}^2+\frac{3}{3}\\ x=\frac{1}{3}{\left(y+3\right)}^2+1\end{array}$

Hence, the standard form of the equation is $x=\frac{1}{3}{\left(y+3\right)}^2+1$

This equation is of the form role="math" localid="1648401713634" $x=a{\left(y-k\right)}^2+h$which represents a parabola.

Hence, the graph of the equation is a parabola.

Graphing the equation:

Step 3. Conclusion.

Hence, the standard form of the equation is$x=\frac{1}{3}{\left(y+3\right)}^2+1$and represents a parabola.