Question

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

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

Short answer

The graph of the equation is

Step-by-step solution

Step 1. Given Information

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

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

Here$a=-2,k=1,h=2$

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

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$(2,1)$

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:

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

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

y-intercept:

Plugging $x=0$in the equation:

role="math" localid="1645432272221" $$\begin{gathered} x=-2{(y-1)}^2+2 \\ 0=-2{(y-1)}^2+2 \\ -2{(y-1)}^2=-2 \\ 2{(y-1)}^2=2 \\ {(y-1)}^2=1 \\ y-1=\pm \sqrt{1} \\ y-1=\pm 1 \\ y=\pm 1+1 \\ y=+1+1,y=-1+1 \\ y=2,y=0 \end{gathered}$$

Hence y-intercepts are $(0,2),(0,0)$

Step 6. Graphing the parabola

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

Conclusion 

The graph of the parabola is