How do you find vertical, horizontal and oblique asymptotes for ((x-1)(x-3))/(x(x-2))?

Jun 9, 2016

vertical asymptotes x = 0 , x = 2
horizontal asymptote y = 1

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve: x(x-2) = 0 → x = 0 , x = 2 are the asymptotes

Horizontal asymptotes occur as

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

now $\frac{\left(x - 1\right) \left(x - 3\right)}{x \left(x - 2\right)} = \frac{{x}^{2} - 4 x + 3}{{x}^{2} - 2 x}$

divide terms on numerator/denominator by ${x}^{2}$

$\frac{{x}^{2} / {x}^{2} - \frac{4 x}{x} ^ 2 + \frac{3}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{2 x}{x} ^ 2} = \frac{1 - \frac{4}{x} + \frac{3}{x} ^ 2}{1 - \frac{2}{x}}$

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

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

Oblique 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 oblique asymptotes.
graph{(x^2-4x+3)/(x^2-2x) [-10, 10, -5, 5]}