How do you find a quadratic function f(x)=ax^2+bx+c for which f(1)=-2, f(-3)=-46, and f(3)=-16?

2 Answers
Jun 2, 2015

I would build a system of three equations in the three unknowns a, b, c and solve it using Cramer's Method:
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Jun 2, 2015

The question gives us 3 simultaneous equations to solve:

-2 = f(1) = a+b+c

-46 = f(-3) = 9a-3b+c

-16 = f(3) = 9a+3b+c

Subtracting a couple of these we find:

30 = -16+46 = f(3) - f(-3)

= (9a+3b+c)-(9a-3b+c) = 6b

So b = 30/6 = 5

Then

-58 = -46-16+4 = f(-3)+f(3)-2f(1)

=(9a-3b+c)+(9a+3b+c)-2(a+b+c)

=16a-2b = 16a-10

Add 10 to both ends to get:

-48 = 16a

So a = -48/16 = -3

Then

-2 = f(1) = a+b+c = -3 + 5 + c = 2 + c#

Subtract 2 from both ends to get:

c = -4

So f(x) = -3x^2+5x-4

Check:

f(1) = -3+5-4 = 5-7 = -2

f(-3) = -3*9-3*5-4 = -27-15-4 = -46

f(3) = -3*9+3*5-4 = -27+15-4 = -16