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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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