Determine the equation of the quadratic function in the form y=ax^2+bx+c that has x-intercepts at -5 and 8 and passes the point (-2,90)

The general form of the quadratic equation is ax^2 + bx + c = 0.

You are being told, in effect, that 2 points on this curve are (-5,0) and (8,0).

Write out the quadratic equation three times, as follows:

a(-5)^2 + b(-5) + c = 0  and  a(8)^2 + b(8) + c = 0.  Both are = to 0 because y is 0 when x is either -5 or +8.  Also,  y = 90 = a(-2)^2 + b(-2) + c = 0, or 
90 = 4a - 2b + c

Then we have 3 equations in 3 unknowns: 
25a - 5b + c = 0
64a + 8b + c = 0
4a - 2b + c = 90

This is a system of linear equations, and there are various ways of solving such a system, including, but not limited to, substitution, matrices, determinants and the like.  I solved this system on my TI-83 Plus calculator using matrix operations, and found that a = -3, b = -9 and c = 120.

Let me know if you need help in solving such a system.

Thus, the quadratic equation in question is

-3x^2 - 9x + 120 = y

Let's check this out!  Does the point (8,0) satisfy  -3x^2 - 9x + 120 = y  ?

Does -3(8)^2 - 9(8) - 120 = 0 ?  Does -3(64) - 72 + 120 = 0 ?  Unfortunately, NO, which means that there is an arithmetic mistake somewhere in my presentation.  Please read thru this work and ask any questions you may have; we will eventually solve the problem correctly.