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

1 Answer
Aug 16, 2016

Answer:

horizontal asymptote at y = 1

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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^2+1=0rArrx^2=-1# which has no real solutions hence there are no vertical asymptotes.

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

as #xto+-oo,f(x)to(1-0)/(1+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^2-3x)/(x^2+1) [-10, 10, -5, 5]}