Question 1037252
Let the quadratic function be {{{f(x) = a(x-r)(x-5)}}}, where x=5 is one of the roots (as given via the x-intercept), and x = r is the other root.

Since y = 10 is the value of the y-intercept, the constant term in the quadratic function must be equal to 5ar, and moreover, 5ar = 10, by hypothesis.

==> ar = 2.

Since the turning point (or the vertex of the parabola) is always midway between the two roots, it follows that {{{z = (r+5)/2}}}.

==> {{{f((r+5)/2) = a((r+5)/2-r)((r+5)/2-5) = -a(r-5)^2/4 = -8}}}, or
{{{a(r-5)^2 = 32}}}.
Divide this equation by the equation ar = 2, giving

{{{(r-5)^2/r = 16}}}, or, 
{{{(r-5)^2 = 16r}}} or
{{{r^2-26r+25 = (r-25)(r-1) = 0}}}, after expanding and simplifying.

==> r = 25 or r = 1.
If r = 25, then a = 2/25.
If r = 1, then a = 2.

Hence there are two possible answers:
(i) The function {{{f(x) = (2/25)(x-25)(x-5)}}} has {{{z = (25+5)/2 = 15}}}, while, 
(ii) The function {{{f(x) = 2(x-1)(x-5)}}} has {{{z = (1+5)/2 = 3}}}.