I don't understand this math problem connected with parabola

Question

a ball is thrown from the ground towards the net at a distance of 4 m. It reaches a height above the net of 3 m. Then it falls again to the ground at a distance of 8 m from the net. Find the maximum height that the ball reaches

Answer 1

3 points on a parabolic path are

(0,0), (8,0) and as it goes over the net (4,3)

(also a 4th point (8,3)

axis of symmetry is x=6

6 is the average of the x coordinates of the two x intercepts (0,0) and (0,8+4)= (0,12)

6 is the average of 0 and 12

it's a downward opening parabola with a<0,

and k>3, greater than height of the net

y =a(x-h)^2 + k

with 3 unknowns

plug in the 3 (or 4) points to get at least 3 equations

solve them for k = maximum point on the parabola

use substitution and elimination, or matrix algebra

y= (-3/4)(x-6)^2 + 27

a=-3/4

h=6

k=27 meters = maximum height

= y coordinate of the vertex

it helps to sketch a rough graph

with the origin and (0,12) as the points at ground level

(6,27)= the maximum point = vertex

with luck there's no mistake in the above calculations

so try the substitution, elimination method and find

a, h, and k, where k= the solution

h(x) = (-3/4)(x-6)^2 + 27

= (-3/4)(x^2 -12x +36) +27

= -3x^2/4 -9x -27+27

= -3x^2/4 +9x

h'(x) = -(6/4)x -9=0

x = 9(4)/6 = 6 = x coordinate of the vertex (maximum point)

h(6) = (-3/4)(6-6)^2 +27 = 27 meters = y coordinate of the vertex = the maximum