Question

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

$x=-3{(y-2)}^2+3$

Short answer

The graph of the equation is

Step-by-step solution

Step 1. Given Information

Given equation of the parabola is$x=-3{(y-2)}^2+3$

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$x=-3{(y-2)}^2+3$

Here$a=-3,k=2,h=3$

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:

$y=2$

Hence the axis of symmetry is$y=2$

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$(3,2)$

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:

$$\begin{gathered} x=-3{(y-2)}^2+3 \\ x=-3{(0-2)}^2+3 \\ x=-3{(-2)}^2+3 \\ x=-3\left(4\right)+3 \\ x=-12+3 \\ x=-9 \end{gathered}$$

Hence the x-intercept is $(-9,0)$

y-intercept:

Plugging $x=0$in the equation:

role="math" localid="1645431300485" $$\begin{gathered} x=-3{(y-2)}^2+3 \\ 0=-3{(y-2)}^2+3 \\ -3{(y-2)}^2=-3 \\ 3{(y-2)}^2=3 \\ {(y-2)}^2=1 \\ y-2=\pm \sqrt{1} \\ y-2=\pm 1 \\ y=\pm 1+2 \\ y=1+2,y=-1+2 \\ y=3,y=1 \end{gathered}$$

Hence y-intercepts are $(0,3),(0,1)$

Step 6. Graphing the parabola

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

Conclusion

The graph of the parabola is