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