## University Calculus: Early Transcendentals (3rd Edition)

- Horizontal asymptote: $y=1$ - Vertical asymptote: $x=-2$ As $x\to\pm\infty$, the dominant term is $1$. As $x\to-2$, the dominant term is $1/(x+2)$.
$$y=\frac{x+3}{x+2}$$ We are interested in the behavior of function $y$ as $x\to\pm\infty$ as well as the behavior of $y$ as $x\to-2$, which is where the denominator is $0$. We can rewrite the function into a polynomial with a remainder as follows: $$y=\frac{x+3}{x+2}=\frac{(x+2)+1}{x+2}=1+\frac{1}{x+2}$$ - As $x\to\pm\infty$, $(x+2)$ approaches $\pm\infty$ as well and $1/(x+2)$ gets closer to $0$. Therefore, the function $y=1+\frac{1}{x+2}$ will approach $1$, meaning that the curve will approach the line $y=1$, which is also the horizontal asymptote. One more thing, because $1/(x+2)$ becomes closer to $0$, $1$ will dominate when x is numerically large. We can say the dominant term is $1$ as $x\to\pm\infty$. - As $x\to-2$, $(x+2)$ will approach $0$ and $1/(x+2)$ will approach $\infty$. Therefore, the function $y=1+\frac{1}{x+2}$ will approach $\infty$, meaning that the curve will approach $\infty$ as well. So $x=-2$ is the vertical asymptote. And since $1/(x+2)$ approaches $\infty$, or we might say becomes infinitely large in absolute value, as $x\to-2$, the dominant term is $1/(x+2)$ here. The graph is shown below. The red curve is the graph of $y=\frac{x+3}{x+2}$, while the blue line is the horizontal asymptote $y=1$ and the green one is the vertical asymptote $x=-2$.