Stitz-Zeager_College_Algebra_e-book

Our solution is 1 1 3 to solve f x g x we look

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Unformatted text preview: x − 6|, we can take points on the graph of g (x) = x2 − x − 6 and apply the absolute value to each of the y values on the parabola. We see from the graph of g that for x ≤ −2 or x ≥ 3, the y values on the parabola are greater than or equal to zero (since the graph is on or above the x-axis), so the absolute value leaves this portion of the graph alone. For x between −2 and 3, however, the y values on the parabola are negative. For example, the point (0, −6) on y = x2 − x − 6 would result in the point (0, | − 6|) = (0, −(−6)) = (0, 6) on the graph of f (x) = |x2 − x − 6|. Proceeding in this manner for all points with x-coordinates between −2 and 3 results in the graph seen above on the right. 2.3 Quadratic Functions 147 y 7 y 7 6 6 5 5 4 4 3 3 2 2 1 1 −3 −2 −1 −1 1 2 3 x −3 −2 −1 −1 −2 −5 −6 x −4 −5 3 −3 −4 2 −2 −3 1 −6 y = g (x) = x2 − x − 6 y = f (x) = |x2 − x − 6| If we take a step back and look at the graphs of...
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