What are the vertical and horizontal asymptotes of #f(x)=5/((x+1)(x-3))#?

Question

Genesis Allen · Accepted Answer

#"vertical asymptotes at "x=-1" and "x=3#
#"horizontal asymptote at "y=0# #"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+1)(x-3)=0# #rArrx=-1" and "x=3" are the 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)=(5/x^2)/(x^2/x^2-(2x)/x^2-3/x^2)=(5/x^2)/(1-2/x-3/x^2)# #"as "xto+-oo,f(x)to0/(1-0-0)# #rArry=0" is the asymptote"# graph{5/((x+1)(x-3)) [-10, 10, -5, 5]}

Eli Assouline · Answer

undefined The vertical asymptotes of the function f(x) = 5/((x+1)(x-3)) are x = -1 and x = 3. There are no horizontal asymptotes for this function.

Jameson Allen · Answer

undefined The vertical asymptotes of the function $ f(x) = \frac{5}{{(x+1)(x-3)}} $ occur where the denominator equals zero. Therefore, the vertical asymptotes are at $ x = -1 $ and $ x = 3 $.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator (0) is less than the degree of the denominator (2), the horizontal asymptote is at $ y = 0 $, which is the x-axis.

So, the vertical asymptotes are at $ x = -1 $ and $ x = 3 $, and the horizontal asymptote is at $ y = 0 $.