The subset of real numbers x for which $\displaystyle f(x) = x^2 -3x + 2$ assumes negative values is an open interval of the form (a,b). Express a and b as integers.

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- Jun 13th 2009, 08:26 PM #1

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- Jun 13th 2009, 08:55 PM #2
Factoring this function, we get f(x) = (x-1) (x-2), a parabola that opens upward and passes through x=1 and x=2. The lower portion of the parabola that is below the horizontal axis (negative values) is the part of the curve for x values between 1 and 2, so the open interval is (1, 2).

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