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

##### 1 Answer

#### Answer:

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_(xto+-oo),f(x)toc" (a constant)"# now

#((x-1)(x-3))/(x(x-2))=(x^2-4x+3)/(x^2-2x)# divide terms on numerator/denominator by

#x^2#

#(x^2/x^2-(4x)/x^2+3/x^2)/(x^2/x^2-(2x)/x^2)=(1-4/x+3/x^2)/(1-2/x)# as

#xto+-oo,f(x)to(1-0+0)/(1-0)#

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]}