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?

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Answer 1 · 2015-06-02T18:16:40

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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Answer 2 · 2015-06-02T18:32:57

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#