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AMC 12 Mock Exam 4 Problem 8

Problem 8 of 25

Let f(x)=x2−6x+cf(x)=x^2-6x+c, where cc is a constant. The smallest value that f(x)f(x) takes as xx ranges over all real numbers is 44. What is f(−1)f(-1)?

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Solution

Completing the square, f(x)=(x−3)2+(c−9)f(x)=(x-3)^2+(c-9), whose minimum value is c−9c-9. So c−9=4c-9=4, c=13c=13, and f(−1)=1+6+13=20f(-1)=1+6+13=\boxed{20}.

Tagged: quadratic minimum (vertex)