Asked by John

Determoine the cubic function with zeros -2, 3, and 4 and F(5) = 28
Answered by Ms Pi_3.14159265358979
Since the zeros are listed... start with them : )
y= a(x+2)(x-3)(x-4)
now you have the F(5) = 28 so set each of your x's to 5 and your y to 28 and solve for "a" : )
Answered by Reiny
So clearly f(x) = a(x+2)(x-3)(x-4)
also f(5) = 28, so
28 = a(7)(2)(1)
a = 2

f(x) = 2(x+2)(x-3)(x-4) , expand it if you need to, I like it better this way
Answered by Schris
@Reiny where'd you get the x-4?
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