How do you find vertical, horizontal and oblique asymptotes for y= x/((x+3)(x-4)y=x(x+3)(x−4)?
1 Answer
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)=0solve (x+3)(x−4)=0
rArrx=-3" and " x=4" are the asymptotes"⇒x=−3 and x=4 are the asymptotes
"Horizontal asymptotes occur as"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]}
