How do you find the asymptotes for # g(x)=(3x^2+2x-1) /( x^2-4)#?
1 Answer
May 31, 2016
vertical asymptotes x = ± 2
horizontal asymptote y = 3
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^2-4=0rArr(x-2)(x+2)=0rArrx=±2#
#rArrx=-2,x=2" are the asymptotes"# Horizontal asymptotes occur as
#lim_(xto+-oo),g(x)to0# divide terms on numerator/denominator by
#x^2#
#((3x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-4/x^2)=(3+2/x-1/x^2)/(1-4/x^2)# as
#xto+-oo,g(x)to(3+0-0)/(1-0)#
#rArry=3" is the asymptote"#
graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}