91Ó°ÊÓ

Given equation of the parabola is $x=-y^2-4y+5$

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 $x=-y^2-4y+5$

Here $a=-1,b=-4,c=5$

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

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

Plugging the values:

$$\begin{gathered} y=-\frac{b}{2a} \\ y=-\frac{(-4)}{2(-1)} \\ y=-\left(\frac{-4}{-2}\right) \\ y=-\left(2\right) \\ y=-2 \end{gathered}$$

Hence the axis of symmetry is$y=-2$

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

Plugging $y=-2$in the given equation:

localid="1645290306471" $$\begin{gathered} x=-y^2-4y+5 \\ x=-{(-2)}^2-4(-2)+5 \\ x=-\left(4\right)+8+5 \\ x=9 \end{gathered}$$

The vertex of parabola is localid="1645290322220" $(9,-2)$

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

The graph of the parabola is