How do you find the vertical, horizontal or slant asymptotes for #(x^2 + x - 2 )/( x^2 + 3x -4)#?

1 Answer
Sep 11, 2016

vertical asymptote at x = - 4
horizontal asymptote at y = 1


The first step is to factorise and simplify f(x).


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 denominator 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)toc" ( a constant)"#

divide terms on numerator/denominator by x


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

#rArry=1" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(x+2)/(x+4) [-10, 10, -5, 5]}