91Ó°ÊÓ

Given equation of the parabola is$x=-3y^2-6y-5$

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

The given equation is$x=-3y^2-6y-5$

Writing the equation in standard form is

$$\begin{gathered} x=-3y^2-6y-5 \\ x=-3y^2-6y-3-2 \\ x=-3(y^2+2y+1)-2 \\ x=-3{(y+1)}^2-2 \end{gathered}$$

Here $a=-3,k=-1,h=-2$

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

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$y=-1$

Since, the vertex and intercepts are only two points, more points are needed to plot the parabola.

Hence assuming few y values and finding the corresponding x values to points can be obtained.

When $y=-2$:

$$\begin{gathered} x=-3{\left(y+1\right)}^2-2 \\ x=-3{\left(-2+1\right)}^2-2 \\ x=-3{\left(-1\right)}^2-2 \\ x=-3\left(1\right)-2 \\ x=-3-2 \\ x=-5 \end{gathered}$$

When $y=1$:

$$\begin{gathered} x=-3{\left(y+1\right)}^2-2 \\ x=-3{\left(1+1\right)}^2-2 \\ x=-3{\left(2\right)}^2-2 \\ x=-3\left(4\right)-2 \\ x=-12-2 \\ x=-14 \end{gathered}$$

Hence the points are$\left(-5,-2\right),\left(-14,1\right)$

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

The graph of the parabola is