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

(a) The given equation has to be determined in standard form.

(b) The graph of the parabola is to be determined using properties.

Step 2. Equation in standard form

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

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

Writing the equation in standard form is

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

Here localid="1645442790284" $a=1,k=-1,h=-4$

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:

$$\begin{gathered} x={(y+1)}^2-4 \\ x={(0+1)}^2-4 \\ x={\left(1\right)}^2-4 \\ x=1-4 \\ x=-3 \end{gathered}$$

Hence the x-intercept is localid="1645537069556" $(-3,0)$

y-intercept:

Plugging $x=0$in the equation:

$$\begin{gathered} x={(y+1)}^2-4 \\ 0={(y+1)}^2-4 \\ {(y+1)}^2=4 \\ (y+1)=\sqrt{4} \\ y+1=\pm 2 \\ y=\pm 2-1 \\ y=2-1,y=-2-1 \\ y=1,y=-3 \end{gathered}$$

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