Answer
Vertex: $\left(-3,-9\right)$
Opening: right
Shape: wider than as $x=y^2$
Horizontal Axis
Work Step by Step
To find the vertex of the given equation, $
x=\dfrac{1}{3}y^2+6y+24
,$ convert the equation in the form $x=a(y-k)^2+h$.
Grouping the $y$-variables together and making the coefficient of $y^2$ equal to $1$, the given equation is equivalent to
\begin{align*}
x&=\left(\dfrac{1}{3}y^2+6y\right)+24
\\\\
x&=\dfrac{1}{3}\left(y^2+18y\right)+24
.\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&=\dfrac{1}{3}\left(y^2+18y+\left(\dfrac{18}{2}\right)^2\right)+\left[24-\dfrac{1}{3}\left(\dfrac{18}{2}\right)^2\right]
\\\\
x&=\dfrac{1}{3}\left(y^2+18y+81\right)+\left[24-27\right]
\\\\
x&=\dfrac{1}{3}\left(y+9\right)^2-3
.\end{align*}
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(-3,-9\right)
$.
In the equation $
x=\dfrac{1}{3}\left(y+9\right)^2-3
,$ the squared variable is $y.$ Thus, the parabola opens left or right. With $a=\dfrac{1}{3}$ (greater than zero), then the parabola opens to the right.
With $|a|=\left|\dfrac{1}{3}\right|=\dfrac{1}{3}$ (between $0$ to $1$), then the shape of the parabola is wider than $x=y^2$.
Since $y$ is the squared variable in the equation, then the parabola has a horizontal axis.
