Answer
Choice D
Work Step by Step
To find the vertex of the given equation, $
x=-y^2-2y+4
,$ convert the equation in the form $x=a(y-k)^2+h$.
Grouping the $x$-variables together and making the coefficient of $x^2$ equal to $1$, the given equation is equivalent to
\begin{align*}
x&=(-y^2-2y)+4
\\
x&=-(y^2+2y)+4
.\end{align*}
Completing the square of the right-side expression by adding $\left(\dfrac{b}{2}\right)^2,$ the equation above is equivalent to
\begin{align*}
x&=-\left(y^2+2y+\left(\dfrac{2}{2}\right)^2\right)+\left[4-(-1)\left(\dfrac{2}{2}\right)^2\right]
\\\\
x&=-\left(y^2+2y+1\right)+\left[4+1\right]
\\
x&=-\left(y+1\right)^2+5
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
(-1)\left(\dfrac{2}{2}\right)^2
$ should be subtracted as well to cancel out the term that was added to complete the square.)
Since the vertex of the equation $x=a(y-k)^2+h$ is given by $(h,k)$, then the vertex of the quadratic equation above is $
\left(5,-1\right)
$. Thus, the vertex is located on the fourth quadrant.
In the equation $
x=-\left(y+1\right)^2+5
,$ the squared variable is $y.$ Thus, the parabola opens left or right. With $a=-1$ (less than zero), then the parabola opens to the left.
With the vertex located at Quadrant $4$ and the parabola opens to the left, then the corresponding graph of the given equation is Choice D.
