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The quartic regression model is p4 x 00144x4 03507x3

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Unformatted text preview: aph of y = f (x) starts and ends above the x-axis. In other words, f (x) is (+) as x → ±∞, and as a result, we no longer need to evaluate f at the test values x = −3 and x = 4. Is there a way to eliminate the need to evaluate f at the other test values? What we would really need to know is how the function behaves near its zeros does it cross through the x-axis at these points, as it does at x = −2 and x = 0, or does it simply touch and rebound like it does at x = 3. From the sign diagram, the graph of f will cross the 188 Polynomial Functions x-axis whenever the signs on either side of the zero switch (like they do at x = −2 and x = 0); it will touch when the signs are the same on either side of the zero (as is the case with x = 3). What we need to determine is the reason behind whether or not the sign change occurs. y (+) 0 (−) 0 (+) 0 (+) −2 0 3 −3 −1 1 x 4 A sketch of y = f (x) Fortunately, f was given to us in factored form: f (x) = x3 (x − 3)2 (x + 2). When we at...
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