How do you find vertical, horizontal and oblique asymptotes for #(sqrt(x^4 + 6x^2 + 9)) / (2x^2 - 10)#?

1 Answer
Mar 21, 2018

Answer:

Vertical asymptotes are #x=sqrt5# and #x=-sqrt5# and horizontal asymptote is #y=1/2#

Explanation:

Vertical asymptotes are given by putting denominator as #0# i.e. #2x^2-10=0#

or #x^2-5=0# i.e. #x=sqrt5# and #x=-sqrt5#

this is so as when #x->+-sqrt5#, #sqrt(x^4+6x^2+9)/(2x^2-10)->+-oo#

For horizontal asymptote, observe that degree of numerator and denominator both is #2# - we take #sqrtx^4# as #x^2# - hence, we have a horizontal asymptote,

as #lim_(x->oo)sqrt(x^4+6x^2+9)/(2x^2-10)#

= #lim_(x->oo)sqrt(1+6/x^2+9/x^4)/(2-10/x^2)#

= #1/2#

Hence, horizontal asymptote is #y=1/2#

There is no oblique or slnting asymptote. For this we should have degree of numerator just one more than that of denominator.

graph{sqrt(x^4+6x^2+9)/(2x^2-10) [-10, 10, -5, 5]}