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.

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!