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**Unformatted text preview: **e. 160 Linear and Quadratic Functions
y
6
5
4
3
2
1
−3 −2 −1
−1 1 2 3 x −2
−3
−4
−5
−6 y = x2 − x − 6 We can see that the graph of f does dip below the x-axis between its two x-intercepts. The zeros
of f are x = −2 and x = 3 in this case and they divide the domain (the x-axis) into three intervals:
(−∞, −2), (−2, 3), and (3, ∞). For every number in (−∞, −2), the graph of f is above the x-axis;
in other words, f (x) > 0 for all x in (−∞, −2). Similarly, f (x) < 0 for all x in (−2, 3), and f (x) > 0
for all x in (3, ∞). We can schematically represent this with the sign diagram below. (+) 0 (−) 0 (+)
−2 3 Here, the (+) above a portion of the number line indicates f (x) > 0 for those values of x; the (−)
indicates f (x) < 0 there. The numbers labeled on the number line are the zeros of f , so we place
0 above them. We see at once that the solution to f (x) < 0 is (−2, 3).
Our next goal is to establish a procedure by which we can generate the sign diagram without
graphing the function. An important property1 of quadratic functions is that if the function is
positive at one point and negative at another, the functi...

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