x^2
____
3x + 1
How do I simplify this to get the Slant Asymptote?
I thought that you couldn't divide x^2 by 3x +1 because their degrees don't match.
Is what "1/3x"? And do you mean 1/(3x) or (1/3)x?
To divide $\displaystyle x^2$ by 3x+1, first write [tex]x^2= x^2+ 0x+ 0.
How many times does "3x" divide into $\displaystyle x^2$? (1/3)x times, of course. But you are not done. Now subtract 3x+1 times (1/3)x from $\displaystyle x^2$: $\displaystyle x^2- (1/3)x(3x+1)= x^2- x^2- (1/3)x= -(1/3)x+ 0$. How many times does 3x divide into -(1/3)x? -1/9 times. And now you subtract (3x+1) times -1/9 from -(1/3)x+ 0: -(1/3)x- (-1/9)(3x+1)= -(1/3)x+ (1/3)x+ 1/9= 1/9.
3x+1 divides into $\displaystyle x^2$ (1/3)x- 1/9 times with remainder 1/9.