University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 108: 66

Answer

- Horizontal asymptote: $y=0$ - Vertical asymptote: $x=3$ As $x\to\pm\infty$, the dominant term is $0$. As $x\to3$, the dominant term is $-3/(x-3)$.
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Work Step by Step

$$y=\frac{-3}{x-3}$$ We are interested in the behavior of function $y$ as $x\to\pm\infty$ as well as the behavior of $y$ as $x\to3$, which is where the denominator is $0$. We can rewrite the function into a polynomial with a remainder as follows: $$y=\frac{-3}{x-3}=0+\frac{-3}{x-3}$$ - As $x\to\pm\infty$, $(x-3)$ approaches $\pm\infty$ as well and $-3/(x-3)$ gets closer to $0$, meaning that the curve will approach the line $y=0$, which is also the horizontal asymptote. One more thing, because $-3/(x-3)$ becomes closer to $0$, we can say the dominant term is $0$ as $x\to\pm\infty$. - As $x\to3$, $(x-3)$ will approach $0$ and $-3/(x-3)$ will approach $-\infty$, meaning that the curve will approach $-\infty$ as well. So $x=3$ is the vertical asymptote. And since $-3/(x-3)$ approaches $-\infty$ as $x\to3$, the dominant term is $-3/(x-3)$. The graph is shown below. The red curve is the graph of $y=-3/(x-3)$, while the blue line is the horizontal asymptote $y=0$ and the green one is the vertical asymptote $x=3$.
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