How do you find the vertical, horizontal and slant asymptotes of: #x/(1-x)^2#?

Question

Madelyn Anderson · Accepted Answer

#"vertical asymptote at " x=1#
#"horizontal asymptote at " y=0# #"let " f(x)=x/(1-x)^2# The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. #"solve "(1-x)^2=0rArrx=1" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by the highest power of x, that is #x^2# #f(x)=(x/x^2)/(1/x^2-(2x)/x^2+x^2/x^2)=(1/x)/(1/x^2-2/x+1)# as #xto+-oo,f(x)to0/(0-0+1)# #rArry=0" is the 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)/((1-x)^2) [-10, 10, -5, 5]}

Benjamin Achtenberg · Answer

undefined To find the vertical asymptotes of the function $ \frac{x}{(1-x)^2} $, we need to determine where the denominator equals zero. In this case, the denominator is $ (1-x)^2 $, which equals zero when $ 1 - x = 0 $. Solving for $ x $, we get $ x = 1 $. Therefore, the vertical asymptote is at $ x = 1 $.

To find the horizontal asymptote, we analyze the behavior of the function as $ x $ approaches positive or negative infinity. Since the degree of the numerator is equal to the degree of the denominator, we divide the leading coefficients of the numerator and denominator to find the horizontal asymptote. In this case, both the numerator and denominator have degree 2, so the horizontal asymptote is given by the ratio of their leading coefficients, which is $ 0/1 = 0 $. Therefore, the horizontal asymptote is at $ y = 0 $.

To find any slant asymptotes, we perform polynomial long division or synthetic division to divide the numerator by the denominator. If the result is a polynomial with a degree greater than or equal to 1, there is a slant asymptote. In this case, when we perform the division, we find that the result is $ 1 + \frac{1}{1-x} $. This expression does not represent a slant asymptote because it's not a polynomial; rather, it's a rational function with a horizontal asymptote. Therefore, there are no slant asymptotes for this function.