Question bank / Nonlinear functions

Math · Advanced MathMedium

The function ff is defined by f(x)=x2+bx+cf(x) = x^2 + bx + c, where bb and cc are constants. The graph of y=f(x)y = f(x) in the xyxy-plane has its vertex at (3,−4)(3, -4). What is the value of cc?

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Answer: 5

The leading coefficient is 11 and the vertex is (3,−4)(3, -4), so f(x)=(x−3)2−4=x2−6x+9−4=x2−6x+5f(x) = (x - 3)^2 - 4 = x^2 - 6x + 9 - 4 = x^2 - 6x + 5. So c=5c = 5.

Answering −4-4 treats the yy-coordinate of the vertex as the constant term, but cc is f(0)f(0), which is 9−4=59 - 4 = 5.

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