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

1 Answer
Sep 11, 2016

Answer:

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

Explanation:

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

#rArrf(x)=((x+2)cancel((x-1)))/((x+4)cancel((x-1)))=(x+2)/(x+4)#

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

#f(x)=(x/x+2/x)/(x/x+4/x)=(1+2/x)/(1+4/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]}