How to find the asymptotes of #f(x) = (2x-4)/(x^2-4)#?

Question

Sophie Adams · Accepted Answer

vertical asymptote x = -2
horizontal asymptote y = 0

The first step here is to factorise and simplify f(x). #(2x-4)/(x^2-4)=(2cancel((x-2)))/(cancel((x-2))(x+2))=2/(x+2)# 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 + 2 = 0 → x = -2 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo),f(x)to0# divide terms on numerator/denominator by x #(2/x)/(x/x+2/x)=(2/x)/(1+2/x)# as #xto+-oo,f(x)to0/(1+0)# #rArry=0" is the asymptote"# graph{2/(x+2) [-10, 10, -5, 5]}

Xavier Alonso · Answer

undefined To find the asymptotes of $ f(x) = \frac{2x - 4}{x^2 - 4} $, you first need to check for vertical asymptotes. Vertical asymptotes occur where the function approaches positive or negative infinity as $ x $ approaches a certain value. Vertical asymptotes exist where the denominator of the function equals zero but the numerator does not.

In this case, the denominator $ x^2 - 4 $ equals zero when $ x = 2 $ or $ x = -2 $. So, $ x = 2 $ and $ x = -2 $ are potential vertical asymptotes.

Next, you should check for horizontal asymptotes. Horizontal asymptotes occur when $ x $ approaches positive or negative infinity, and the function approaches a constant value. For rational functions like this one, horizontal asymptotes can be determined by comparing the degrees of the numerator and denominator.

In $ f(x) = \frac{2x - 4}{x^2 - 4} $, the degree of the numerator is 1, and the degree of the denominator is 2. Since the degree of the denominator is greater than the degree of the numerator, there is a horizontal asymptote at $ y = 0 $, which is the $ x $-axis.

So, the vertical asymptotes are $ x = 2 $ and $ x = -2 $, and the horizontal asymptote is $ y = 0 $.