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What are the asymptotes of #f(x)=(x^2+1)/(x^2-9)#?

Answer 1

#"vertical asymptotes at "x=+-3#
#"horizontal asymptote at "y=1#

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^2-9=0rArr(x-3)(x+3)=0#

#rArrx=-3" and "x=3" 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+1/x^2)/(x^2/x^2-9/x^2)=(1+1/x^2)/(1-9/x^2)#

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

#rArry=1" is the asymptote"# graph{(x^2+1)/(x^2-9) [-10, 10, -5, 5]}

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Answer 2

Scarlett Allison

The asymptotes of f(x)=(x^2+1)/(x^2-9) are vertical asymptotes at x = 3 and x = -3, and there are no horizontal asymptotes.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.