How do you find the asymptotes for #y=1/(2-x)#?

Question

Chloe Alexander · Accepted Answer

vertical asymptote at x = 2?
horizontal asymptote at y = 0 vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero. solve : 2 - x = 0 → x = 2 is the equation. horizontal asymptotes occur as # lim_(x→±∞) f(x) → 0 # If the degree of the numerator is less than the degree 0f the denominator then the equation is y = 0. here the degree of numerator < degree of denominator and so equation of horizontal asymptote is y = 0 here is the graph of the function as an illustration of them. graph{1/(2-x) [-10, 10, -5, 5]}

Cora Alexander · Answer

undefined To find the asymptotes for the function $ y = \frac{1}{2 - x} $, we first identify any vertical asymptotes by determining where the denominator becomes zero. In this case, the denominator $ 2 - x $ becomes zero when $ x = 2 $. So, there is a vertical asymptote at $ x = 2 $.

Next, we check for horizontal or slant asymptotes. For rational functions like this one, if the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at $ y = 0 $. In this case, the degree of the numerator is less than the degree of the denominator (which is 1), so there is a horizontal asymptote at $ y = 0 $.

Therefore, the asymptotes for the function $ y = \frac{1}{2 - x} $ are:

1. Vertical asymptote: $ x = 2 $
2. Horizontal asymptote: $ y = 0 $