How do you find the Vertical, Horizontal, and Oblique Asymptote given #y = (8 x^2 + x - 2)/(x^2 + x - 72)#?

Question

Cora Alexander · Accepted Answer

The vertical asymptotes are #x=-8# and #x=9#
No oblique asymptote
The horizontal asymptote is #y=8# Let's factorise the denominator #x^2+x-72=(x+8)(x-9)# So, #y=(8x^2+x-2)/(x^2+x-72)=(8x^2+x-2)/((x+8)(x-9))# The domain of #y# is #D_y=RR-{-8,+9}# As we cannot divide by #0#, #x!=-8# and #x!=9# The vertical asymptotes are #x=-8# and #x=9# As the degree of the numerator is #=# to the degree of the denominator, there are no oblique asymptotes. To calculate the limits of #x->+-oo#, we take the terms of highest degree in the numerator and the denominator. #lim_(x->+-oo)y=lim_(x->+-oo)(8x^2)/x^2=8# So, the horizontal asymptote is #y=8# graph{(y-(8x^2+x-2)/(x^2+x-72))(y-8)=0 [-58.5, 58.55, -29.3, 29.23]}

Cora Alexander · Answer

undefined To find the vertical asymptotes of a rational function, set the denominator equal to zero and solve for $x$.

$$ x^2 + x - 72 = 0 $$

Factor the quadratic equation:

$$ (x - 8)(x + 9) = 0 $$

The vertical asymptotes occur where the denominator equals zero, so $x = 8$ and $x = -9$ are the vertical asymptotes.

To find horizontal asymptotes, compare the degrees of the numerator and denominator:

- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$.
- If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients.
- If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

In this case, both the numerator and denominator have the same degree (2). So, the horizontal asymptote is the ratio of the leading coefficients:

$$ y = \frac{8}{1} = 8 $$

To find oblique asymptotes (if they exist), divide the numerator by the denominator using polynomial long division or synthetic division. If the quotient approaches a linear function as $x$ approaches positive or negative infinity, then that linear function is the oblique asymptote.

In this case, perform polynomial long division or synthetic division to divide $8x^2 + x - 2$ by $x^2 + x - 72$. Then, analyze the quotient to see if it approaches a linear function as $x$ approaches positive or negative infinity. If it does, that linear function is the oblique asymptote.