How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x) = ((x – 10)(x + 5) )/( (13x + 10)(10–x))#?

Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x) =  ((x – 10)(x + 5) )/( (13x + 10)(10–x))#?

Liam Brennan · Accepted Answer

The vertical asymptote is #x=-10/13#. The horizontal asymptote is #y=-1/13#. (There's a removable discontinuity at #x=10#.)

First notice that #10-x=-(x-10)#. So we can simplify the given function: #f(x) = ((x – 10)(x + 5) )/( (13x + 10)(10–x)) = -frac(x+5)(13x+10),x!=10#. (Since that factor reduces out we know that the function has a removable discontinuity--or hole--at x = 10, but that wasn't asked.) Now, vertical asymptotes are the zeros of the factors of the denominator that do not cancel with factors of the numerator, so in this case a vertical asymptote is at #x=-10/13#. Horizontal asymptotes are the ratio of the leading coefficients of the numerator and denominator if both the numerator and denominator have the same degree (which they do in this case). The horizontal asymptote in this case is #y=-1/13# for this function because the leading coefficient in the numerator is 1, the leading coefficient in the denominator is 13, and there's a negative just hanging out from the cancelled factor.

Madelyn Anderson · Answer

undefined To find the vertical asymptotes of $ f(x) = \frac{(x - 10)(x + 5)}{(13x + 10)(10 - x)} $, we need to determine where the denominator equals zero and the numerator does not. The vertical asymptotes occur at the values of x that make the denominator zero.

Setting the denominator $ (13x + 10)(10 - x) $ equal to zero, we find:

$$ 13x + 10 = 0 \Rightarrow x = -\frac{10}{13} $$
$$ 10 - x = 0 \Rightarrow x = 10 $$

So, the vertical asymptotes are at $ x = -\frac{10}{13} $ and $ x = 10 $.

To find horizontal and oblique asymptotes, we look at the behavior of the function as $ x $ approaches positive and negative infinity.

As $ x $ approaches positive or negative infinity, the terms involving $ x $ in the numerator and denominator dominate the function. In this case, the degree of the numerator and denominator are the same (both are quadratic), so we divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote.

$$ \lim_{x \to \infty} \frac{(x - 10)(x + 5)}{(13x + 10)(10 - x)} = \lim_{x \to \infty} \frac{x^2}{-13x^2} = \lim_{x \to \infty} \frac{-1}{13} = -\frac{1}{13} $$

Therefore, the horizontal asymptote is $ y = -\frac{1}{13} $.

Since the degree of the numerator is less than the degree of the denominator, there is no oblique asymptote.