# How do you find the asymptotes for f(x)=(x^2-2x)/(x^2-5x+4)?

Feb 19, 2016

vertical asymptotes at x = 1 and x = 4
horizontal asymptote at y = 1

#### Explanation:

Vertical asymptotes occur when the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

solve ${x}^{2} - 5 x + 4 = 0$
factoring gives:
(x-1)(x-4) = 0$\Rightarrow x = 1 , x = 4$

Horizontal asymptotes occur as lim_(x→±∞) f(x) → 0

If the degree of the numerator and denominator are equal , then the equation can be found by taking the ratio of leading coefficients.
Here they are both of degree 2 hence:

$y = \frac{1}{1} = 1 \text{ is the equation}$

here is the graph of the function to illustrate these.
graph{(x^2-2x)/(x^2-5x+4) [-20, 20, -10, 10]}