Question

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

$x=4{(y+1)}^2-4$

Short answer

The graph of the equation is

Step-by-step solution

Step 1. Given Information

Given equation of the parabola is$x=4{(y+1)}^2-4$

The graph of the parabola is to be determined

Step 2. Equation of the parabola

The equation of a parabola in standard form $x=a{(y-k)}^2+h$

The given equation is $x=4{(y+1)}^2-4$

Here $a=4,k=-1,h=-4$

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

Step 3. Axis of symmetry

The axis of symmetry of the parabola $x=a{(y-k)}^2+h$is$y=k$

Plugging the values:

$y=-1$

Hence the axis of symmetry is$y=-1$

Step 4. Vertex of a parabola

The vertex of the parabola $x=a{(y-k)}^2+h$is$(h,k)$

The vertex of the given parabola is$(-4,-1)$

Step 5. Intercepts

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:

$$\begin{gathered} x=4{(y+1)}^2-4 \\ x=4{(0+1)}^2-4 \\ x=4{\left(1\right)}^2-4 \\ x=4-4 \\ x=0 \end{gathered}$$

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

y-intercept:

Plugging $x=0$in the equation:

localid="1645433676926" $$\begin{gathered} x=4{(y+1)}^2-4 \\ 0=4{(y+1)}^2-4 \\ 4{(y+1)}^2=4 \\ {(y+1)}^2=1 \\ y+1=\pm \sqrt{1} \\ y=\pm \sqrt{1}-1 \\ y=\pm 1-1 \\ y=1-1,y=-1-1 \\ y=0,y=-2 \end{gathered}$$

Hence the y-intercepts arelocalid="1645536930764" $(0,0),(0,-2)$

Step 6. Graphing the parabola

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

Conclusion 

The graph of the parabola is