How do you find the asymptotes for #f(x)=(4x)/(x^2-8x+16)#?

Question

James Allen · Accepted Answer

HA= #y= 0#
VA= x= +/- 4
No slant asymptote

factored form of #x^2-8x+16# is #(x-4) (x+4)# so therefor the VA are +/- 4 no slant asymptote because the degree of the numerator is not one more higher than the degree of the denominator. Ha is 0 because the degree of x is higher on the bottom than the top.

Madelyn Anderson · Answer

undefined To find the asymptotes for $ f(x) = \frac{4x}{x^2 - 8x + 16} $, you first need to examine the behavior of the function as $ x $ approaches certain values.

1. **Vertical Asymptotes:** Vertical asymptotes occur when the denominator of the function becomes zero, but the numerator does not. So, find the values of $ x $ that make the denominator zero:

$$ x^2 - 8x + 16 = 0 $$

This quadratic equation can be factored into $(x - 4)^2 = 0$, which yields a repeated root of $ x = 4 $. Therefore, there is a vertical asymptote at $ x = 4 $.

2. **Horizontal Asymptotes:** Horizontal asymptotes occur when $ x $ approaches positive or negative infinity. To find these, examine the behavior of the function as $ x $ goes to positive or negative infinity.

$$ \lim_{{x \to \infty}} \frac{4x}{x^2 - 8x + 16} = 0 $$
$$ \lim_{{x \to -\infty}} \frac{4x}{x^2 - 8x + 16} = 0 $$

Both limits approach zero. Therefore, there are horizontal asymptotes at $ y = 0 $.

So, the asymptotes for $ f(x) = \frac{4x}{x^2 - 8x + 16} $ are:
- Vertical asymptote: $ x = 4 $
- Horizontal asymptote: $ y = 0 $