Question

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

$x=-y^2-2y+3$

Short answer

The graph of the equation is

Step-by-step solution

Step 1. Given Information

Given equation of the parabola is $x=-y^2-2y+3$

The graph of the parabola is to be determined.

Step 2. Equation of the parabola

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

The given equation is $x=-y^2-2y+3$

Here $a=-1,b=-2,c=3$

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

Step 3. Axis of symmetry

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

Plugging the values:

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

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

Step 4. Vertex of a parabola

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

Plugging $y=-1$in the given equation:

$$\begin{gathered} x=-y^2-2y+3 \\ x=-{(-1)}^2-2(-1)+3 \\ x=-\left(1\right)+2+3 \\ x=4 \end{gathered}$$

The vertex of the 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:

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

Hence the x-intercept is localid="1645266465700" $(3,0)$

y-intercept:

Plugging $x=0$in the equation:

$$\begin{gathered} x=-y^2-2y+3 \\ 0=-y^2-2y+3 \\ {y}^2+2y-3=0 \end{gathered}$$

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

$$\begin{gathered} y^2+2y-3=0 \\ {y}^2-1y+3y-3=0 \\ y(y-1)+3(y-1)=0 \\ (y-1)(y+3)=0 \\ y-1=0,y+3=0 \\ y=1,y=-3 \end{gathered}$$

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

Step 6. Graphing the parabola

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

Conclusion

The graph of the parabola is