How to find the asymptotes of #f(x) = (x+1) / (x^2 +3x - 4)# ?
1 Answer
Jun 20, 2016
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" and " 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 the variable , 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"#
graph{(x+1)/(x^2+3x-4) [-10, 10, -5, 5]}