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Stitz-Zeager_College_Algebra_e-book

# The solutions to f x g x are precisely the x values

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Unformatted text preview: lue function. Example 2.3.4. Graph f (x) = |x2 − x − 6|. Solution. Using the deﬁnition of absolute value, Deﬁnition 2.4, we have f (x) = − x2 − x − 6 , if x2 − x − 6 < 0 x2 − x − 6, if x2 − x − 6 ≥ 0 The trouble is that we have yet to develop any analytic techniques to solve nonlinear inequalities such as x2 − x − 6 < 0. You won’t have to wait long; this is one of the main topics of Section 2.4. Nevertheless, we can attack this problem graphically. To that end, we graph y = g (x) = x2 − x − 6 using the intercepts and the vertex. To ﬁnd the x-intercepts, we solve x2 − x − 6 = 0. Factoring gives (x − 3)(x + 2) = 0 so x = −2 or x = 3. Hence, (−2, 0) and (3, 0) are x-intercepts. The y −1 intercept is found by setting x = 0, (0, −6). To ﬁnd the vertex, we ﬁnd x = − 2ba = − 2(1) = 1 , and 2 2 y = 1 − 1 − 6 = − 25 = −6.25. Plotting, we get the parabola seen below on the left. To obtain 2 2 4 points on the graph of y = f (x) = |x2 −...
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