*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **ote, the graph of f nevertheless crosses the x-axis
at (0, 0). The myth that graphs of rational functions can’t cross their horizontal asymptotes is
completely false, as we shall see again in our next example.
Example 4.2.2. Sketch a detailed graph of g (x) = 2x2 − 3x − 5
.
x2 − x − 6 Solution.
1. Setting x2 − x − 6 = 0 gives x = −2 and x = 3. Our domain is (−∞, −2) ∪ (−2, 3) ∪ (3, ∞).
2. Factoring g (x) gives g (x) = (2x−5)(x+1)
(x−3)(x+2) . There is no cancellation, so g (x) is in lowest terms. 3. To ﬁnd the x-intercept we set y = g (x) = 0. Using the factored form of g (x) above, we ﬁnd
the zeros to be the solutions of (2x − 5)(x + 1) = 0. We obtain x = 5 and x = −1. Since
2
both of these numbers are in the domain of g , we have two x-intercepts, 5 , 0 and (−1, 0).
2
To ﬁnd the y -intercept, we set x = 0 and ﬁnd y = g (0) = 5 , so our y -intercept is 0, 5 .
6
6
4. Since g (x) was given to us in lowest terms, we have, once again by Theorem 4.1 vertical
x−5)(x+1)
asymptotes x = −2 and x = 3....

View
Full
Document