How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)=(6x^2+2x-1 )/ (x^2-1)#?

Question

Ariana Alvarez · Accepted Answer

The vertical asymptotes are #x=1# and #x=-1#
The horizontal asymptote is #y=6#
No oblique asymptote

The denominator is #(x^2-1)=(x-1)(x+1)# As we cannot divde by #0#, #x!=+-1# Therefore the vertical asymptotes are #x=1# and #x=-1# As the degree of the numerator #=# degree of the denominator, we don't expect a slant asymptote. For the limit, we take the term with the highest coefficient #lim_(x->+-oo)f(x)=lim_(x->+-oo)(6x^2)/x^2=6# The horizontal asymptote is #y=6# graph{(y-(6x^2+2x-1)/(x^2-1))(y-6)=0 [-12.83, 12.49, -1.84, 10.83]}

Liam Brennan · Answer

undefined To find the vertical asymptotes of a rational function $ f(x) = \frac{{6x^2 + 2x - 1}}{{x^2 - 1}} $:

1. Set the denominator equal to zero and solve for $ x $.
2. Any value of $ x $ that makes the denominator zero is a vertical asymptote.

To find the horizontal asymptotes:

1. Compare the degrees of the numerator and denominator.
2. If the degree of the numerator is greater, there is no horizontal asymptote.
3. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $.
4. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find the oblique asymptote:

1. If the degree of the numerator is exactly one more than the degree of the denominator, there is an oblique (slant) asymptote.
2. Use polynomial long division or synthetic division to divide the numerator by the denominator.
3. The quotient obtained is the equation of the oblique asymptote.