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

1 Answer
Jan 1, 2017

Answer:

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

Explanation:

The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : #(x-1)(x-3)=0rArrx=1" or " x=3#

#rArrx=1" and " x=3" are the asymptotes"#

Horizontal asymptotes occur as

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

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

divide terms on numerator/denominator by the highest power of x, that is #x^2#

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

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

#rArry=1" is the asymptote"#

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+1)^2)/((x-1)(x-3)) [-20, 20, -10, 10]}