91Ó°ÊÓ

Given equation of the parabola is $x=y^2+6y+8$

The graph of the parabola is to be determined.

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

The given equation is $x=y^2+6y+8$

Here $a=1,b=6,c=8$

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

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

Plugging the values:

$$\begin{gathered} y=-\frac{b}{2a} \\ y=-\frac{6}{2\left(1\right)} \\ y=-\frac{6}{2} \\ y=-3 \end{gathered}$$

Hence the axis of symmetry is$y=-3$

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

Plugging $y=-3$in the given equation:

$$\begin{gathered} x=y^2+6y+8 \\ x={(-3)}^2+6(-3)+8 \\ x=9-18+8 \\ x=-1 \end{gathered}$$

The vertex of the parabola is $(-1,-3)$

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:

$$\begin{gathered} x=y^2+6y+8 \\ x={\left(0\right)}^2+6\left(0\right)+8 \\ x=0+0+8 \\ x=8 \end{gathered}$$

Hence the x-intercept is$(8,0)$

y-intercept:

Plugging $x=0$in the equation:

role="math" localid="1645266715095" $$\begin{gathered} x=y^2+6y+8 \\ 0=y^2+6y+8 \\ {y}^2+6y+8=0 \end{gathered}$$

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

role="math" localid="1645266858613" $$\begin{gathered} y^2+6y+8=0 \\ {y}^2+4y+2y+8=0 \\ y(y+4)+2(y+4)=0 \\ (y+2)(y+4)=0 \\ (y+2)=0,(y+4)=0 \\ y=-2,y=-4 \end{gathered}$$

Hence the y-intercepts are $(0,-2),(0,-4)$

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

The graph of the parabola is