Question

Graph the rational function (x^2-8x+12)/(x-6)

Answer 1

Explanation:

`f(x)=(x^(2) - 8 x + 12)/(x - 6)`

VERTICAL ASYMPTOTES

The line x = L is a vertical asymptote of the function
`y=(x^(2) - 8 x + 12)/(x - 6)`

, if the limit of the function (one-sided) at this point is infinite.

In other words, it means that possible points are points where the denominator equals 0 or doesn't exist.

So, find the points where the denominator equals 0 and check them.

x=6, check:

`\lim_(x \to 6^+)\left(x - 2\right)=4`

Since the limit is finite, check another limit.

`\lim_(x \to 6^-)\left((x^(2) - 8 x + 12)/(x - 6)\right)=4`

Since the limit is finite, then x=6 is not a vertical asymptote.

HORIZONTAL ASYMPTOTES

Line y=L is a horizontal asymptote of the function y=f(x), if either
`\lim_(x \to \infty) f{\left(x \right)}=L`

or
`\lim_(x \to -\infty) f{\left(x \right)}=L` , and L is finite.

Calculate the limits:

`\lim_(x \to \infty)\left((x^(2) - 8 x + 12)/(x - 6)\right)=\infty`

`\lim_(x \to -\infty)\left((x^(2) - 8 x + 12)/(x - 6)\right)=-\infty`

Thus, there are no horizontal asymptotes.

SLANT ASYMPTOTES

Do polynomial long division
`(x^(2) - 8 x + 12)/(x - 6)=x - 2` Thus, the slant asymptote is y=x−2.

Answer

No vertical asymptotes.

No horizontal asymptotes.

Slant asymptote: y=x−2