How do you find the vertical, horizontal or slant asymptotes for #y=(x+3)/(x-3)#?

Question

Lily Anderson · Accepted Answer

vertical asymptote x = 3
horizontal asymptote y = 1

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero. solve : x - 3 = 0 → x = 3 is the asymptote Horizontal asymptotes occur as #lim_(x to +- oo), y to 0 # divide terms on numerator/denominator by x #(x/x +3/x)/(x/x-3/x)=(1+3/x)/(1-3/x)# as #x to +- oo , y to (1+0)/(1-0)# # rArr y = 1 " is the asymptote "# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes. graph{(x+3)/(x-3) [-10, 10, -5, 5]}

Madelyn Anderson · Answer

undefined To find the vertical asymptote of the function $ y = \frac{x+3}{x-3} $, we need to look for values of $ x $ that make the denominator equal to zero but do not make the numerator equal to zero. In this case, the denominator $ x - 3 $ equals zero when $ x = 3 $. Since the numerator $ x + 3 $ does not equal zero at $ x = 3 $, we have a vertical asymptote at $ x = 3 $.

To find the horizontal asymptote, we examine the behavior of the function as $ x $ approaches positive or negative infinity. As $ x $ approaches infinity, the terms with the highest power dominate the behavior of the function. In this case, both the numerator and the denominator have the same highest power, which is 1. Therefore, we divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote. So, the horizontal asymptote is $ y = 1 $.

There is no slant asymptote for this function because the degree of the numerator is not greater than the degree of the denominator. If the degree of the numerator were greater than the degree of the denominator by exactly 1, then there would be a slant asymptote.