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

$x=-4y^2$

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

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 $x=-4y^2$

Here $a=-4,b=0,c=0$

Since a is negative, the parabola opens to the left.

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

Plugging the values:

$$\begin{gathered} y=-\frac{b}{2a} \\ y=-\frac{0}{2\left(-4\right)} \\ y=0 \end{gathered}$$

Hence the axis of symmetry is$y=0$

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

Plugging $y=0$in the given equation:

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

The vertex of the parabola is$(0,0)$

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:

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

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

y-intercept:

Plugging $x=0$in the equation:

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

Hence the y-intercept is $(0,0)$

Since, the vertex and intercepts are the same point, more point are needed to plot the parabola.

Hence assuming few y values and finding the corresponding x values to points can be obtained.

When $y=-1$:

role="math" localid="1645254329276" $$\begin{gathered} x=-4y^2 \\ x=-4{(-1)}^2 \\ x=-4\left(1\right) \\ x=-4 \end{gathered}$$

Whenrole="math" localid="1645254441704" $y=1$:

role="math" localid="1645254350930" $$\begin{gathered} x=-4y^2 \\ x=-4{\left(1\right)}^2 \\ x=-4\left(1\right) \\ x=-4 \end{gathered}$$

Hence the points are$(-4,-1),(-4,1)$

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

The graph of the parabola is