How do you find the horizontal asymptote for #(2x^2)/(x^2-4)#?

Question

Lincoln Anderson · Accepted Answer

vertical asymptotes x = ± 2
horizontal asymptote y = 2

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero. solve: #x^2-4=0rArr(x-2)(x+2)=0rArrx=±2# #rArrx=-2,x=2" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by #x^2# #((2x^2)/x^2)/(x^2/x^2-4/x^2)=2/(1-4/x^2)# as #xto+-oo,f(x)to2/(1-0)# #rArry=2" is the asymptote"# graph{(2x^2)/(x^2-4) [-10, 10, -5, 5]}

Lily Anderson · Answer

undefined To find the horizontal asymptote of the function $ \frac{2x^2}{x^2 - 4} $, compare the degrees of the numerator and denominator polynomials. If the degree of the numerator is equal to or less than the degree of the denominator, the horizontal asymptote is $ y = 0 $. If the degree of the numerator is greater than the degree of the denominator by exactly one, there is a slant asymptote. If the degree of the numerator is greater than the degree of the denominator by more than one, there is no horizontal asymptote.

In this case, both the numerator and denominator have a degree of 2. To find the horizontal asymptote, divide the leading coefficients of the numerator and denominator. The horizontal asymptote is the equation $ y = \frac{a}{b} $, where $ a $ is the leading coefficient of the numerator and $ b $ is the leading coefficient of the denominator. Therefore, the horizontal asymptote for $ \frac{2x^2}{x^2 - 4} $ is $ y = \frac{2}{1} = 2 $.