Question

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

Answer 1

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 - 5x + 4 = 0`
factoring gives:
(x-1)(x-4) = 0`rArr 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 = 1/1 = 1 " 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]}