# How do you find the vertical, horizontal and slant asymptotes of: #f(x)=(2x+4)/(x^2-3x-4)#?

##### 1 Answer

vertical asymptotes x = -1 , x = 4

horizontal asymptote y = 0

#### Explanation:

The denominator of f(x) cannot be zero as this is 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^2-3x-4=0rArr(x-4)(x+1)=0#

#rArrx=-1" and " x=4" 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#

#((2x)/x^2+4/x^2)/(x^2/x^2-(3x)/x^2-4/x^2)=(2/x+4/x^2)/(1-3/x-4/x^2)# as

#xto+-oo,f(x)to(0+0)/(1-0-0)#

#rArry=0" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1, denominator-degree 2)Hence there are no slant asymptotes.

graph{(2x+4)/(x^2-3x-4) [-10, 10, -5, 5]}