How do you identify all horizontal and slant asymptote for #f(x)=(3x^2+1)/(x^2+x+9)#?

2 Answers
Dec 18, 2016

Answer:

Horizontal: #larr y = 3 rarr#

Explanation:

By actual division,

#f(x) = 3-(3x+26) / (x^2+x+9)#

y = quotient = 3 gives the asymptote.

For x <= -26/3, y>=3#.

On the left side, (-10, 3.04040404..)# is on the curve.

Yet, as #x to -oo, y to 3#, turning back to the asymptote

graph{y(x^2+x+9)-3x^2-1 = 0 [-10, 10, -5.21, 5.21]}

Dec 18, 2016

Answer:

horizontal asymptote at y = 3

Explanation:

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" ( a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=((3x^2)/x^2+1/x^2)/(x^2/x^2+x/x^2+9/x^2)=(3+1/x^2)/(1+1/x+9/x^2)#

as #xto+-oo,f(x)to(3+0)/(1+0+0)#

#rArry=3" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2) Hence there are no slant asymptotes.
graph{(3x^2+1)/(x^2+x+9) [-10, 10, -5, 5]}