How do you find vertical, horizontal and oblique asymptotes for #(x^2-2x)/(x+1)#?

Let's do a long division

So,

Therefore

graph{(y-(x^2-x)/(x+1))(y-x)=0 [-36.53, 36.57, -18.26, 18.27]}

To find the vertical, horizontal, and oblique asymptotes for the function $\frac{x^2 - 2x}{x + 1}$, follow these steps:

1. Vertical Asymptotes: Vertical asymptotes occur where the denominator of the rational function becomes zero but the numerator does not. In this case, the vertical asymptote occurs when $x + 1 = 0$, so $x = -1$ is the vertical asymptote.

2. Horizontal Asymptotes: To find horizontal asymptotes, compare the degrees of the numerator and denominator of the rational function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

In this case, both the numerator and denominator have a degree of 1. Therefore, to find the horizontal asymptote, divide the leading coefficient of the numerator by the leading coefficient of the denominator.

$\lim_{x \to \infty} \frac{x^2 - 2x}{x + 1} = \lim_{x \to \infty} \frac{x^2/x - 2x/x}{x/x + 1/x} = \lim_{x \to \infty} \frac{x - 2}{1 + 1/x} = \frac{\infty - 0}{1 + 0} = \infty$

So, there is no horizontal asymptote.

3. Oblique Asymptotes: If the degree of the numerator is exactly one greater than the degree of the denominator, the graph will have an oblique asymptote. To find the oblique asymptote, perform polynomial long division or use the method of synthetic division to divide the numerator by the denominator.

$\frac{x^2 - 2x}{x + 1} = x - 3 + \frac{-3}{x + 1}$

As $x$ approaches infinity, $\frac{-3}{x + 1}$ approaches 0. Therefore, the oblique asymptote is $y = x - 3$.