The minimum value of a quadratic function appears as a constant in the vertex form of its equation, which can be found from the standard form by completing the square. Rewriting
f(x)=(x+6)(x−4) in standard form gives
f(x)=x2+2x−24. Since the coefficient of the linearterm is
2, the equation for
f(x) can be rewritten in terms of
(x+1)2 as follows:
f(x)=x2+2x−24=(x2+2x+1)−1−24=(x+1)2−25
Since the square of a real number is always nonnegative, the vertex form
f(x)=(x+1)2−25 shows that the minimum value of f is
−25 (and occurs at
x=−1). Therefore, this equivalent form of f shows the minimum value of
f as a constant.
Choices A and C are incorrect because they are not equivalent to the given equation for
f. Choice B is incorrect because the minimum value of
f, which is
−25, does not appear as a constant or a coefficient.