Question

which statement is true concerning the vertex and axis of symmetry of $h(x) = -2x^2 + 8x$? the vertex is at $(2, 2)$ and the axis of symmetry is $y = 2$. the vertex is at $(2, 8)$ and the axis of symmetry is $x = 2$. the vertex is at $(0, 0)$ and the axis of symmetry is $y = 2$. the vertex is at $(0, 0)$ and the axis of symmetry is $x = 2$.

Explanation:

Step1: Find axis of symmetry

For quadratic $ax^2+bx+c$, axis is $x=-\frac{b}{2a}$.
Here $a=-2$, $b=8$, so $x=-\frac{8}{2(-2)} = 2$.

Step2: Find vertex y-coordinate

Substitute $x=2$ into $h(x)$:
$h(2) = -2(2)^2 + 8(2) = -8 + 16 = 8$.
So vertex is $(2, 8)$.

Answer:

The vertex is at (2, 8) and the axis of symmetry is $x = 2$.