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College Algebra, 2013a

College Algebra, 2013a

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330 Rational Functions<br />

ˆ The behavior of y = h(x) as x →∞: If x →∞,then 3<br />

x+2<br />

≈ very small (+). This means<br />

h(x) ≈ 2x − 1 + very small (+), or that the graph of y = h(x) is a little bit above the<br />

line y =2x − 1asx →∞.<br />

Graphically we have<br />

4<br />

3<br />

2<br />

1<br />

y<br />

−1<br />

x<br />

−2<br />

−3<br />

−4<br />

6. To make our sign diagram, we place an ‘”’ abovex = −2 andx = −1 and a ‘0’ above x = − 1 2 .<br />

On our four test intervals, we find h(x) is (+) on (−2, −1) and ( − 1 2 , ∞) and h(x) is(−) on<br />

(−∞, −2) and ( −1, − 1 2)<br />

. Putting all of our work together yields the graph below.<br />

(−) ” (+) ” (−) 0 (+)<br />

−2 −1 − 1 2<br />

9<br />

8<br />

7<br />

6<br />

5<br />

4<br />

3<br />

2<br />

1<br />

−4 −3 −1 −1 1 2 3 4<br />

−2<br />

−3<br />

−4<br />

−5<br />

−6<br />

−7<br />

−8<br />

−9<br />

−10<br />

−11<br />

−12<br />

−13<br />

−14<br />

y<br />

x<br />

We could ask whether the graph of y = h(x) crosses its slant asymptote. From the formula<br />

h(x) =2x − 1+ 3<br />

x+2 , x ≠ −1, we see that if h(x) =2x − 1, we would have 3<br />

x+2<br />

= 0. Since this will<br />

never happen, we conclude the graph never crosses its slant asymptote. 14<br />

14 But rest assured, some graphs do!

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