How do you find the vertical, horizontal or slant asymptotes for #(2x)/(x-1)#?

Question

Elias Ackerman · Accepted Answer

vertical asymptote x = 1
horizontal asymptote y = 2

When a rational function's denominator approaches zero, vertical asymptotes occur. Set the denominator to zero to find the equation. The asymptote is x = 1. Solve this: x - 1 = 0. Horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0 # Divide the numerator/denominator by x to find the equation. #(2x)/(x-1) = ((2x)/x)/(x/x- 1/x) = 2/(1 - 1/x)# As x tends to ∞ , #1/x → 0 rArr y = 2/1 = 2 " is the asymptote "# This is the function's graph: graph{2x/(x-1) [-10, 10, -5, 5]}

Sadie Ackerman · Answer

undefined To find the vertical asymptote, set the denominator equal to zero and solve for $x$. In this case, $x - 1 = 0$ gives $x = 1$. Therefore, there is a vertical asymptote at $x = 1$.

To find horizontal asymptotes, compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$. If the degree of the numerator equals the degree of the denominator, divide the leading coefficients. In this case, both the numerator and the denominator have a degree of 1. So, the horizontal asymptote is $y = 2/1 = 2$.

There are no slant asymptotes since the degree of the numerator is not greater than the degree of the denominator by 1 or more.