Expert: Ahmed Salami - 10/4/2004

Question
A certain quadratic function has the form f(x)+ ax^2 + bx + c, where a, b, and c not equal to zero.If f(-2)=-18, f(0)=-8, f(b)=c, determine algebraically the values of a,b,and c.The answer is: a=-1,b=3,c=-8, but I can't seem to get it. Please explain.Thank you.

Answer
Hi jeff,
f(x)= ax^2 + bx + c
f(-2)= a(-2)^2 + b(-2) + c = 4a - 2b + c = -18
f(0)= a(0)^2 + b(0) + c = c = -8
we now know that c = -8 already
f(b)= ab^2 + b(b) + c
then,
ab^2 + b^2 + c = c since f(b)= c
ab^2 + b^2 = 0
b^2(a + 1)= 0
There actually two pairs of values of a and b that would satisfy our requirements. For the equation
b^2(a + 1)= 0, its either
b^2 = 0 which gives b = 0
or a + 1 = 0 which gives a = -1
Now, with b = 0, the equation
4a - 2b + c = -18 becomes
4a - 2(0)+ (-8)= -18
4a - 8 = -18
4a = -10
a = -10/4 = -2.5
Also, with a = -1, the equation
4a - 2b + c = -18 becomes
4(-1)- 2b + (-8)= -18
-4 - 2b - 8 = -18
-2b - 12 = -18
-2b = -6
b = 3
Since the question requires that none of a, b, c should be zero, we stick to the second solution a = -1, b = 3, c = -8
I hope i have helped.
Regards.