How do you find the asymptotes for #f(x) =(x^2-1)/(x^2+4)#?

Question

Madelyn Anderson · Accepted Answer

horizontal asymptote at y = 1

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes. solve: #x^2+4=0rArrx^2=-4# This has no real solutions hence there are no vertical asymptotes. Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by the highest power of x that is #x^2# #f(x)=(x^2/x^2-1/x^2)/(x^2/x^2+4/x^2)=(1-1/x^2)/(1+4/x^2)# as #xto+-oo,f(x)to(1-0)/(1+0)# #rArry=1" is the asymptote"# graph{(x^2-1)/(x^2+4) [-10, 10, -5, 5]}

Benjamin Achtenberg · Answer

undefined To find the asymptotes for the function $ f(x) = \frac{x^2 - 1}{x^2 + 4} $, first, determine the vertical asymptotes by identifying the values of $ x $ that make the denominator zero. Set $ x^2 + 4 = 0 $ and solve for $ x $. There are no real solutions, so there are no vertical asymptotes.

To find the horizontal asymptotes, compare the degrees of the numerator and the denominator. Since both have the same degree (2), divide the leading coefficients. The horizontal asymptote is $ y = \frac{1}{1} = 1 $.

So, the function has a horizontal asymptote at $ y = 1 $, and there are no vertical asymptotes.