How do you find vertical, horizontal and oblique asymptotes for #y= x/((x+3)(x-4)#?
1 Answer
May 3, 2017
Explanation:
The denominator of y cannot be zero as this would make y undefined. Equating the denominator yo zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
#"solve " (x+3)(x-4)=0#
#rArrx=-3" and " x=4" are the asymptotes"#
#"Horizontal asymptotes occur as"#
#lim_(xto+-oo),ytoc" (a constant)"#
#"divide terms on numerator/denominator by the highest power"#
#"of x, that is " x^2#
#y=(x/x^2)/(x^2/x^2-x/x^2-12/x^2)=(1/x)/(1-1/x-12/x^2)# as
#xto+-oo,yto0/(1-0-0)#
#rArry=0" is the asymptote"#
graph{x/((x+3)(x-4)) [-10, 10, -5, 5]}