# Thread: VA,HA & Slant Asymptotes

1. ## VA,HA & Slant Asymptotes

I was wondering if this function had a vertical and/or horizontal and/or slant asymptotes?

Function:

f(x)= (x^3-2x+5)/(3x^2+4x-1)

I know how to find the slant asymptotes with long division, but i can't seem to do it with this function.

Thanks

2. Originally Posted by dto

I know how to find the slant asymptotes with long division, but i can't seem to do it with this function.
Use $\displaystyle (x^3{\color{red}+0x^2}-2x+50)\div (3x^2+4x-1)$

There will plenty of fractions to keep you busy.

Originally Posted by dto
I was wondering if this function had a vertical and/or horizontal and/or slant asymptotes?

Will have vertical asymptotes at $\displaystyle 3x^2+4x-1=0$

You can solve this with the quadratic formula.

3. so i think the answer for the slant is: y= 1/3x-4/9 R1/9x+4/9)/(1/3x-4/9)

and is that suppose to be a 50 or a 5?

Thanks

4. Originally Posted by dto
so i think the answer for the slant is: y= 1/3x-4/9 R1/9x+4/9)/(1/3x-4/9)

and is that suppose to be a 50 or a 5?

Thanks
It's supposed to be a 5!

5. so is the slant y=1/3x-4/9?

6. Originally Posted by dto
so is the slant y=1/3x-4/9?

Thats what I get!

7. and for the va, is it (-2+/- square root of 7)/ 3?

i know it looks funky, but I don't know how to do the symbols.
If it is like that, how do i graph the va?

Thanks

p.s. about the slant, sweet! what I did wrong was, I did not realize that it was going to be fractions! thanks alot!

8. yet another question! 4x/x-2.... if the bottom is more then the top, that means there is no slant right?

Correct me if I am wrong:
HA= 4
VA=2
Slant= 0 or should it be none?