In this section we develop a technique for investigating the range of values assumed by a quadratic function. In subsequent work we shall assume as known the results of the examples in Section 10.1 and of Problems 1 to 8 in Section 10.1.

Example: Let f(x) = ax² + bx + c, where a, b, and c are real numbers and Let D be the discriminant b² - 4ac. Show that if D > 0, then f(x) takes on both positive and negative values.

Solution: Completing squares, we obtain

If x = -b/2a, 2ax + b = 0, and so f(-b/2a) = -D/4a. We first consider the case in which a > 0. This and D > 0 imply that f(x) = -D/4a < 0. We wish to show that f(x) also takes on positive values. We consider values of x greater that  Then

Thus we have proved the desired result for the case a > 0. If a < 0, let g(x) = -ax² - bx - ac. Since the coefficient of x² in g(x) is positive, g(x) takes on both positive and negative values by the previous case. Then so does f(x) = -g(x).
 

Monday, June 22, 1998