How do you find vertical, horizontal and oblique asymptotes for #1/(x^2-2x-8)#?

Question

Caleb Alexander · Accepted Answer

The vertical asymptotes are #x=4# and #x=-2#
The horizontal asymptote is #y=0#
No oblique asymptote

Let's factorise the denominator #x^2-2x-8=(x-4)(x+2)# As you canot divide by #0#, #=>#, #x!=4# and #x!=-2# The vertical asymptotes are #x=4# and #x=-2# As the degree of the numerator is #<# than the degree of the denominator, there is no oblique asymptote. #lim_(x->+-oo)1/(x^2-2x-8)=lim_(x->+-oo)1/x^2=0^+# The horizontal asymptote is #y=0# graph{(y-1/(x^2-2x-8))(y-10000(x-4))(y-1000(x+2))(y)=0 [-3.85, 7.25, -2.893, 2.66]}

James Allen · Answer

undefined To find the vertical, horizontal, and oblique asymptotes for the function $f(x) = \frac{1}{x^2 - 2x - 8}$, follow these steps:

1. **Vertical Asymptotes**:
   Vertical asymptotes occur where the denominator of the function becomes zero, but the numerator does not. Set the denominator equal to zero and solve for $x$. Any $x$ value obtained is a potential vertical asymptote.

For $f(x) = \frac{1}{x^2 - 2x - 8}$, the denominator $x^2 - 2x - 8$ factors into $(x - 4)(x + 2)$. Setting each factor equal to zero, we find that $x = 4$ and $x = -2$ are vertical asymptotes.

2. **Horizontal Asymptotes**:
   Horizontal asymptotes occur when $x$ approaches positive or negative infinity. To find horizontal asymptotes, analyze the behavior of the function as $x$ approaches positive and negative infinity.

As $x$ approaches positive or negative infinity, the terms with the highest degree in the numerator and denominator dominate the function. Since both the numerator and denominator are quadratic, divide both by $x^2$ and take the limit as $x$ approaches infinity.

$$ \lim_{x \to \pm \infty} \frac{1}{x^2 - 2x - 8} = 0 $$

This means the horizontal asymptote is $y = 0$.

3. **Oblique Asymptotes**:
   Oblique asymptotes occur when the degree of the numerator is exactly one greater than the degree of the denominator. To find the oblique asymptote, perform polynomial long division or synthetic division to divide the numerator by the denominator.

$$ \frac{1}{x^2 - 2x - 8} = \frac{0}{x - 4} + \frac{1}{x + 2} $$

As $x$ approaches positive or negative infinity, the term $\frac{0}{x - 4}$ tends to zero, leaving the oblique asymptote as $y = \frac{1}{x + 2}$.