How do you find the vertical, horizontal or slant asymptotes for #f(x) = (x+1) / (x^2 +3x - 4)#?

1 Answer
Jun 27, 2016

Answer:

vertical asymptotes x = -4 , x = 1
horizontal asymptote y = 0

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve: #x^2+3x-4=0rArr(x+4)(x-1)=0#

#rArrx=-4,x=1" 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 , in this case #x^2#

#(x/x^2+1/x^2)/(x^2/x^2+(3x)/x^2-4/x^2)=(1/x+1/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{(x+1)/(x^2+3x-4) [-10, 10, -5, 5]}