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

Question

Serenity Jones · Accepted Answer

vertical asymptotes x = 1 , x = 3
horizontal asymptote y = 1

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]}

Theodore Amidon · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y=0. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. For slant asymptotes, divide the numerator by the denominator using polynomial long division, and the resulting quotient is the equation of the slant asymptote.