How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= 1/(x-2)#?

Question

Chloe Alexander · Accepted Answer

vertical asymptote x = 2
horizontal asymptote y = 0

The denominator of f(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. solve: x - 2 = 0 → x = 2 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by x #(1/x)/(x/x-2/x)=(1/x)/(1-2/x)# as #xto+-oo,f(x)to0/(1-0)# #rArry=0" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0 , denominator-degree 1 ) Hence there are no oblique asymptotes. graph{(1)/(x-2) [-10, 10, -5, 5]}

Cora Alexander · Answer

undefined To find the vertical asymptote of the function $f(x) = \frac{1}{x - 2}$, you set the denominator equal to zero and solve for $x$. In this case, $x - 2 = 0$, so $x = 2$. Therefore, the vertical asymptote is $x = 2$.

To find the horizontal asymptote, we examine the behavior of the function as $x$ approaches positive and negative infinity. As $x$ becomes large in either direction, the term $\frac{1}{x - 2}$ approaches zero. Hence, there is a horizontal asymptote at $y = 0$.

There is no oblique (slant) asymptote for this function because the degree of the numerator is less than the degree of the denominator. When this condition is met, there is no oblique asymptote; only vertical and horizontal asymptotes apply. Therefore, in this case, there is no oblique asymptote.