# How do you find the vertical, horizontal or slant asymptotes for #y= (x+1)^2 / ((x-1)(x-3))#?

##### 1 Answer

vertical asymptotes x = 1 , x = 3

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-1)(x-3) = 0 → x = 1 , x = 3 are the asymptotes.

Horizontal asymptotes occur as

#lim_(xto+-oo) , y to 0# When the degree of the numerator and denominator are equal , as is the case here (both of degree 2) then the equation is the ratio of leading coefficients.

#y=(x+1)^2/((x-1)(x-3))=(x^2+2x+1)/(x^2-4x+3)# The leading coefficients(coefficient of

#x^2# terms) are

#y=1/1=1" is the equation of asymptote "# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.

graph{(x^2+2x+1)/(x^2-4x+3) [-14.24, 14.24, -7.12, 7.12]}