Parabolas

Given that

$\begin{aligned} x & = - 2y^2 + 12y - 15 \\ & = 3 - 2(y^2 - 6y +9) \\ x & = 3 - 2(y-3)^2 \\ 3-x & = 2(y-3)^2 \end{aligned}$

The parabola is of the form $X = aY^2$ , where $X=3-x$ , $Y= y-3$ , and $a=2$ . The vertex is where $X=0$ and $Y=0$ or $3-x = 0$ and $y-3 = 0$ , or $\boxed{(3,3)}$ .

Since the y-coordinate are all the same in the options, let's plug $y=3$ back to the equation.

$x=-2(3)^2+12(3)-15=3$ , hence the vertex is $(3,3)$ .

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Here is how we derive the y-coordinate:

$x=-2y^2+12y-15=-2(y-3)^2+3$ , hence $y-3=0$

By definition, vertex is that point where the equation representing the parabola has repeated roots, I. e., the discriminant of the equation is zero . So in the given problem, we have $6^2=4\times 1\times \dfrac {15+x}{2}\implies x=3$ from this.

Then $y=\dfrac {6}{2}=3$ , and the vertex is at $\boxed {(3,3)}$ .