How do you find the vertical, horizontal and slant asymptotes of: y = (x)/(x^2+4)y=xx2+4?
1 Answer
horizontal asymptote at y = 0
Explanation:
The denominator of y cannot be zero as this would make y undefined. Equating the denominator to 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^2+4=0rArrx^2=-4x2+4=0⇒x2=−4 This has no real solutions hence there are no vertical 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+4/x^2)=(1/x)/(1+4/x^2) as
xto+-oo,yto0/(1+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)/(x^2+4) [-10, 10, -5, 5]}
