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Stitz-Zeager_College_Algebra_e-book

# In a similar fashion we recall that to test an

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Unformatted text preview: h satisfy the equation y = f (x). That is, the point (x, y ) is on the graph of f if and only if y = f (x). Example 1.7.1. Graph f (x) = x2 − x − 6. Solution. To graph f , we graph the equation y = f (x). To this end, we use the techniques outlined in Section 1.3. Speciﬁcally, we check for intercepts, test for symmetry, and plot additional points as needed. To ﬁnd the x-intercepts, we set y = 0. Since y = f (x), this means f (x) = 0. f (x) 0 0 x−3=0 x = = = or = x2 − x − 6 x2 − x − 6 (x − 3)(x + 2) factor x+2=0 −2, 3 So we get (−2, 0) and (3, 0) as x-intercepts. To ﬁnd the y -intercept, we set x = 0. Using function notation, this is the same as ﬁnding f (0) and f (0) = 02 − 0 − 6 = −6. Thus the y -intercept is (0, −6). As far as symmetry is concerned, we can tell from the intercepts that the graph possesses none of the three symmetries discussed thus far. (You should verify this.) We can make a table analogous to the ones we made in Section 1.3, plot the points and connect the dots in a somewhat plea...
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