Please help me with this quadratics question Thanks so much

Please help me with this quadratics question. Thanks so much!

Let f(x) = 8x - 2x^2. Part of the graph of f is shown below. Find the x-intercepts of the graph. Write down the equation of the axis of symmetry, Find the y-coordinate of the vertex.

Solution

f(x) = -2x^2 +8x

y = -2x^2 +8x

x intercept : plug y =0

-2x^2 +8x =0 ; x( -2x +8) =0

x =0 ; x= 4

In vertex form the equation : y = -2x^2 +8x

= -2(x^2 -4x)

= -2(x^2 -4x +4 -4) = -2(x -2)^2 + 8

y = -2(x -2)^2 + 8  

So, vertex : ( 2, 8)

y coordinate of vertex y =8

axis of symmetry : x= 2