SOLUTION: Find the horizontal or oblique asymptotes of g(x) F(x)= 9x^3+2x/3x^2+1 I know you need to use long divison but I am stuck

Question 1033529: Find the horizontal or oblique asymptotes of g(x)
F(x)= 9x^3+2x/3x^2+1
I know you need to use long divison but I am stuck
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
equation is y = (9x^3 + 2x) / (3x^2 + 1)

if the degree of the numerator is less than or equal to the degree of the denominator, than you will get a horizontal asymptote.

if the degree of the numerator is greater than the degree of the denominator by 1 degree, then you will get a slant asymptote.

the degree of the numerator is one more than the degree of the denominator so you have a slant asymptote.

divide the denominator into the numerator and you will get y = 3x plus a remainder.

forget the remainder and the equation of your slant asymptote is y = 3x.

here's the graph of your equation and the equation of the slant asymptote.

$$$

the graph of the horizontal or slant asymptote CAN cross the graph of the equation.

that is exactly what it does here.

it crosses the graph of the equation at x = 0.

here's the same graph blown up so you can see the intersection point more clearly.

$$$

here's a reference on horizontal and slant asymptotes.

http://www.coolmath.com/precalculus-review-calculus-intro/precalculus-algebra/18-rational-functions-finding-horizontal-slant-asymptotes-01

you, of course, need to know how to do polynomial division.

if you have a problem with that, look here:

http://www.purplemath.com/modules/polydiv2.htm

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