How do you find the vertical, horizontal or slant asymptotes for # f(x) = (3x) /( x+4)#?

Question

Serenity Johnson · Accepted Answer

#"vertical asymptote at "x=-4#
#"horizontal asymptote at "y=3# The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. #"solve "x+4=0rArrx=-4" is the asymptote"# #"horizontal asymptotes occur as"# #lim_(xto+-oo),f(x)to c" ( a constant)"# #"divide terms on numerator/denominator by x"# #f(x)=((3x)/x)/(x/x+4/x)=3/(1+4/x)# #"as "xto+-oo,f(x)to3/(1+0)# #y=3" is the asymptote"# Slant asymptotes occur when the degree of the denominator is greater than the degree of the denominator. This is not the case here hence there are no slant asymptotes. graph{(3x)/(x+4) [-20, 20, -10, 10]}

James Allen · Answer

undefined To find the vertical asymptote, set the denominator equal to zero and solve for $ x $. In this case, $ x + 4 = 0 $, so $ x = -4 $ is the vertical asymptote.

To find horizontal asymptotes, compare the degrees of the numerator and denominator. Since both have degree 1, divide the leading coefficient of the numerator by the leading coefficient of the denominator. In this case, $ \frac{3}{1} = 3 $. So, the horizontal asymptote is $ y = 3 $.

There are no slant asymptotes for rational functions where the degree of the numerator is less than the degree of the denominator. Therefore, there are no slant asymptotes for this function.