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

Question

James Allen · Accepted Answer

The vertical asymptote is #x=1#
No slant asymptote.
The horizontal asymptote is #y=0# The domain of #f(x)# is #D_f(x)=RR-{1}# As you cannot divide by #0#, #x!=1# The vertical asymptote is #x=1# The degree of the numerator is #<# than the degree of the denominator, there is no slant asymptote. To calculate the limits as #x->+-oo#, we take the termsof highest degree in the numerator and the denominator #lim_(x->-oo)f(x)=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^(-)# #lim_(x->+oo)f(x)=lim_(x->+oo)x/x^2=lim_(x->+oo)1/x=0^(+)# The horizontal asymptote is #y=0# graph{x/(x-1)^2 [-10, 10, -5, 5]}

Christian Antunez · Answer

undefined To find the vertical asymptotes of the function $f(x) = \frac{x}{(x-1)^2}$, we look for values of $x$ that make the denominator zero but not the numerator. In this case, the vertical asymptote occurs at $x = 1$.

To find the horizontal asymptote, we compare the degrees of the numerator and denominator. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is located at $y = 0$.

To find the slant (oblique) asymptote, we divide the numerator by the denominator using polynomial long division. After performing the division, if the quotient is a polynomial function, it represents the slant asymptote. In this case, the quotient is $y = 1$. Therefore, the slant asymptote is $y = x + 1$.