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

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

Explanation:

We have the function
`f(x) = (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^+) (x^2+6x-8)/(x-8) = +\infty`

and

`\lim_(x\rightarrow 8^-) (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) (x^2+6x-8)/(x-8) = L\in\mathbb{R}`

But, actually

`\lim_(x\rightarrow \infty) (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) = (f(x))/(x) = \lim_(x\rightarrow +\infty) (x^2+6x-8)/(x(x-8)) = (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) (x^2+6x-8)/(x-8) -x = \lim_(x\rightarrow +\infty) (14x-8)/(x-8) = 14`

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