Answer
Choice E
Work Step by Step
To find the vertex of the given equation, $
x=3y^2+6y+5
,$ 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&=(3y^2+6y)+5
\\
x&=3(y^2+2y)+5
.\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&=3\left(y^2+2y+\left(\dfrac{2}{2}\right)^2\right)+\left[5-3\left(\dfrac{2}{2}\right)^2\right]
\\\\
x&=3\left(y^2+2y+1\right)+\left[5-3\right]
\\
x&=3\left(y+1\right)^2+2
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
3\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(2,-1\right)
$. Thus, the vertex is located on the fourth quadrant.
In the equation $
x=3\left(y+1\right)^2+2
,$ the squared variable is $y.$ Thus, the parabola opens left or right. With $a=3$ (greater than zero), then the parabola opens to the right.
With the vertex located at Quadrant $4$ and the parabola opens to the right, then the corresponding graph of the given equation is Choice E.
