How do you find the Vertical, Horizontal, and Oblique Asymptote given #(x^2+1)/(x^2-1)#?
1 Answer
vertical asymptotes at x = ± 1
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^2-1=0rArr(x-1)(x+1)=0rArrx=+-1#
#rArrx=-1" and " x=1" are the asymptotes"# Horizontal asymptotes occur as
#lim_(xto+-oo),f(x)toc" (a constant)"# divide numerator/denominator by the highest power of x, that is
#x^2#
#f(x)=(x^2/x^2+1/x^2)/(x^2/x^2-1/x^2)=(1+1/x^2)/(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+1)/(x^2-1) [-10, 10, -5, 5]}
