Answer:
There is no horizontal asymptote for the function f(x).
Step-by-step explanation:
We have the function [tex]f(x) = \frac{x^2+6x-8}{x-8}[/tex]. Notice that if [tex]x=8[/tex] the function is not defined, because the denominator of the fractions equals zero, and the numerator don't. This fact is equivalent to the existence of a vertical asymptote at [tex]x=8[/tex]. In mathematical language:
[tex]\lim_{x\rightarrow 8^+} \frac{x^2+6x-8}{x-8} = +\infty[/tex]
and
[tex]\lim_{x\rightarrow 8^-} \frac{x^2+6x-8}{x-8} = -\infty.[/tex]
Now, in case that f(x) has an horizontal asymptote the following must hold:
[tex]\lim_{x\rightarrow +\infty} \frac{x^2+6x-8}{x-8} = L\in\mathbb{R}[/tex]
But, actually
[tex]\lim_{x\rightarrow \infty} \frac{x^2+6x-8}{x-8} = \infty.[/tex]
Hence, there is no horizontal asymptote.
Anyway, f(x) has an asymptote, but no horizontal. In order to obtain the slope of the asymptote, we need to find the following limit:
[tex]\lim_{x\rightarrow +\infty} = \frac{f(x)}{x} = \lim_{x\rightarrow +\infty} \frac{x^2+6x-8}{x(x-8)} = \frac{x^2+6x-8}{x^2-8x} = 1.[/tex]
Then asymptote has equation [tex]y=x+n[/tex]. To find [tex]n[/tex] we calculate the limit
[tex]\lim_{x\rightarrow +\infty} (f(x)-mx) = \lim_{x\rightarrow +\infty} \frac{x^2+6x-8}{x-8} -x = \lim_{x\rightarrow +\infty} \frac{14x-8}{x-8} = 14[/tex]
Hence, the asymptote at [tex]+\infty[/tex] is [tex] y = x+14[/tex].
