How do you find the vertical, horizontal and slant asymptotes of: ( x^2-2x)/(x^2-5x+4)?

Dec 8, 2016

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

Explanation:

The denominator of the function f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : ${x}^{2} - 5 x + 4 = 0 \Rightarrow \left(x - 1\right) \left(x - 4\right) = 0$

$\Rightarrow x = 1 \text{ and " x=4" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} - \frac{2 x}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{5 x}{x} ^ 2 + \frac{4}{x} ^ 2} = \frac{1 - \frac{2}{x}}{1 - \frac{5}{x} + \frac{4}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{1 - 0 + 0}$

$\Rightarrow y = 1 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2 ) Hence there are no slant asymptotes.
graph{(x^2-2x)/(x^2-5x+4) [-10, 10, -5, 5]}