Question

State the horizontal asymptote of the rational function.f(x) = quantity x squared plus six x minus eight divided by quantity x minus eight

Answer 1

f(x)=(x²+6x-8)/(x-8) has a vertical asymptote at x=8. When x is large f(x) approaches x which is a 45 degree line (gradient 1). There is no horizontal asymptote.

Answer 2

Answer:

There is no horizontal asymptote for the function f(x).

Step-by-step explanation:

We have the function $f(x) = \frac{x^2+6x-8}{x-8}$. Notice that if $x=8$ 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 $x=8$. In mathematical language:

$\lim_{x\rightarrow 8^+} \frac{x^2+6x-8}{x-8} = +\infty$

and

$\lim_{x\rightarrow 8^-} \frac{x^2+6x-8}{x-8} = -\infty.$

Now, in case that f(x) has an horizontal asymptote the following must hold:

$\lim_{x\rightarrow +\infty} \frac{x^2+6x-8}{x-8} = L\in\mathbb{R}$

But, actually

$\lim_{x\rightarrow \infty} \frac{x^2+6x-8}{x-8} = \infty.$

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:

$\lim_{x\rightarrow +\infty} = \frac{f(x)}{x} = \lim_{x\rightarrow +\infty} \frac{x^2+6x-8}{x(x-8)} = \frac{x^2+6x-8}{x^2-8x} = 1.$

Then asymptote has equation $y=x+n$. To find $n$ we calculate the limit

$\lim_{x\rightarrow +\infty} (f(x)-mx) = \lim_{x\rightarrow +\infty} \frac{x^2+6x-8}{x-8} -x = \lim_{x\rightarrow +\infty} \frac{14x-8}{x-8} = 14$

Hence, the asymptote at $+\infty$ is $y = x+14$.