Respuesta :
Answer:
Step-by-step explanation:
[tex]f(x)=\frac{x^{2} - 8 x + 12}{x - 6}[/tex]
VERTICAL ASYMPTOTES
The line x = L is a vertical asymptote of the function [tex]y=\frac{x^{2} - 8 x + 12}{x - 6}[/tex]
, 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:
[tex]\lim_{x \to 6^+}\left(x - 2\right)=4[/tex]
Since the limit is finite, check another limit.
[tex]\lim_{x \to 6^-}\left(\frac{x^{2} - 8 x + 12}{x - 6}\right)=4[/tex]
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 [tex]\lim_{x \to \infty} f{\left(x \right)}=L[/tex]
or [tex]\lim_{x \to -\infty} f{\left(x \right)}=L[/tex] , and L is finite.
Calculate the limits:
[tex]\lim_{x \to \infty}\left(\frac{x^{2} - 8 x + 12}{x - 6}\right)=\infty[/tex]
[tex]\lim_{x \to -\infty}\left(\frac{x^{2} - 8 x + 12}{x - 6}\right)=-\infty[/tex]
Thus, there are no horizontal asymptotes.
SLANT ASYMPTOTES
Do polynomial long division [tex]\frac{x^{2} - 8 x + 12}{x - 6}=x - 2[/tex] Thus, the slant asymptote is y=x−2.
Answer
No vertical asymptotes.
No horizontal asymptotes.
Slant asymptote: y=x−2
