Home > Precalculus > Asymptotes

What are the asymptotes for #x/(x^2+x-6)#?

Answer 1

Serenity Johnson

Refer to explanation

We see that as #x->+-oo# , #x/(x^2+x-6)->0# hence #y=0# horizontal asymptote.

and as #x->2# , #x/(x^2+x-6)->oo# and

#x->-3# , #x/(x^2+x-6)->oo#

so vertical asymptote are #x=2,x=-3#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Sadie Ackerman

The asymptotes for the function ( \frac{x}{x^2 + x - 6} ) can be found by analyzing the behavior of the function as ( x ) approaches positive or negative infinity.

Horizontal Asymptote: As ( x ) approaches positive or negative infinity, the function approaches zero. Therefore, the horizontal asymptote is ( y = 0 ).
Vertical Asymptotes: Vertical asymptotes occur where the denominator of the function becomes zero. Factorizing the denominator, ( x^2 + x - 6 ), we get ( (x + 3)(x - 2) ). Hence, vertical asymptotes occur at ( x = -3 ) and ( x = 2 ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.