How do you find the Vertical, Horizontal, and Oblique Asymptote given #(x^2+4)/(6x-5x^2)#?
1 Answer
Explanation:
#"let " f(x)=(x^2+4)/(6x-5x^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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes
#"solve " 6x-5x^2=0rArrx(6-5x)=0#
#rArrx=0" and " x=6/5" are the 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+4/x^2)/((6x)/x^2-(5x^2)/x^2)=(1+4/x^2)/(6/x-5)# as
#xto+-oo,f(x)to(1+0)/(0-5)#
#rArry=-1/5" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 2 ) Hence there are no oblique asymptotes.
graph{(x^2+4)/(6x-5x^2) [-10, 10, -5, 5]}
