h=-16(1.25)^2+40(1.25)+50

Question

Trying to find the y in the vertex x is 1.25 (1.25,y).

Answer 1

Let y=-16x²+40x+50,

y=50-16(x²-5x/2+25/16-25/16),

y=50-16((x-5/4)²-25/16),

y=50-16(x-5/4)²+25,

y=75-16(x-5/4)².

The axis of symmetry is x=5/4=1.25, a vertical line (parallel to the y-axis), not a single point.

The single point is in fact the vertex of the parabola at (1.25,75).