Unformatted text preview: To make
this exclusion speciﬁc, we write h(x) = (2x+1)(x+1) , x = −1.
3. To ﬁnd the x-intercepts, as usual, we set h(x) = 0 and solve. Solving (2x+1)(x+1) = 0 yields
x = − 1 and x = −1. The latter isn’t in the domain of h, so we exclude it. Our only x2
intercept is − 2 , 0 . To ﬁnd the y -intercept, we set x = 0. Since 0 = −1, we can use the
reduced formula for h(x) and we get h(0) = 1 for a y -intercept of 0, 1 .
4. From Theorem 4.1, we know that since x = −2 still poses a threat in the denominator of
the reduced function, we have a vertical asymptote there. As for x = −1, we note the factor
(x + 1) was canceled from the denominator when we reduced h(x), and so it no longer causes
trouble there. This means we get a hole when x = −1. To ﬁnd the y -coordinate of the hole,
we substitute x = −1 into (2x+1)(x+1) , per Theorem 4.1 and get 0. Hence, we have a hole on
12 Bet you never thought you’d never see that stuﬀ again before the Final Exam! 4.2 Graphs of Rational Functio...
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