In the following exercises, graph each equation by using properties.

$x=y^2-4y-12$

Given equation of the parabola is $x=y^2-4y-12$

The graph of the parabola is to be determined.

The equation of a parabola in general form is$x=a{y}^2+by+c$

The given equation is $x=y^2-4y-12$

Here $a=1,b=-4,c=-12$

Since a is positive, the parabola opens to the right.

The axis of symmetry of the parabola $x=a{y}^2+by+c$is $y=-\frac{b}{2a}$

Plugging the values:

$$\begin{gathered} y=-\frac{b}{2a} \\ y=-\frac{(-4)}{2\left(1\right)} \\ y=-(-2) \\ y=2 \end{gathered}$$

Hence the axis of symmetry is$y=2$

The vertex of the parabola $x=a{y}^2+by+c$can be determined by plugging and solving for x.

Plugging $y=2$in the given equation:

$$\begin{gathered} x=y^2-4y-12 \\ x={\left(2\right)}^2-4\left(2\right)-12 \\ x=4-8-12 \\ x=-16 \end{gathered}$$

The vertex of the parabola is $(-16,2)$

x-intercept is the point where the graph crosses the x-axis and y is 0.

y-intercept is the point where the graph crosses the y-axis and x is 0.

x-intercept:

Plugging $y=0$in the equation:

$$\begin{gathered} x=y^2-4y-12 \\ x={\left(0\right)}^2-4\left(0\right)-12 \\ x=0-0-12 \\ x=-12 \end{gathered}$$

Hence the x-intercept is $(-12,0)$

y-intercept:

Plugging $x=0$in the equation :

$$\begin{gathered} x=y^2-4y-12 \\ 0=y^2-4y-12 \\ {y}^2-4y-12=0 \end{gathered}$$

The quadratic equation has to be solved to get y values:

localid="1645318830248" $$\begin{gathered} y^2-4y-12=0 \\ {y}^2+2y-6y-12=0 \\ y(y+2)-6(y+2)=0 \\ (y+2)(y-6)=0 \\ (y+2)=0,(y-6)=0 \\ y=-2,y=6 \end{gathered}$$

Hence the y-intercepts are localid="1645318844250" $(0,-2),(0,6)$

Using the properties, plotting the points and graphing the parabola:

The graph of the parabola is