# Homework Help: Vertical asymptote

1. Sep 25, 2007

### teffy3001

its asking me to find the value of p so that the function won't have a vertical asymptote:

g(x) = (3x + p) / (x^3 + 8)

however, i'm not sure how to figure that out...any help?

2. Sep 25, 2007

### genneth

For what value of x is the vertical asymptote? What value should the numerator take there to get rid of it?

3. Sep 25, 2007

### HallsofIvy

Do you understand what a " vertical asymptote" IS and why a graph of this function might have one? Answering this will involve factoring the denominator.

4. Sep 25, 2007

### teffy3001

after i factor the denominator though...where do i go from there?

5. Sep 25, 2007

### rock.freak667

What values of x would make the entire denominator be equal to zero??

..........................................

and hence make the function tend to infinity?

6. Feb 29, 2008

### teffy3001

i know if i set x equal to 2 that the denominator will equal 0 but then the point is not to have a vertical asymptote so how do i find the value of p?

7. Feb 29, 2008

### Dick

If the numerator is zero at x=2, then you might not have a vertical asymptote.

8. Feb 29, 2008

### symbolipoint

WHAT?
Look at the denominator. If x=-2, then x^3 + 8 = 0. The denominator makes the function undefined at x=-2, so this is the vertical asymtote.

9. Feb 29, 2008

### Dick

Sure. Sorry. x=-2. I was only looking at the OP's last post.

10. Feb 29, 2008

### HallsofIvy

You still might not have a vertical asymptote at x= -2!

Yes, at x= -2, the denominator goes to 0. As long as the numerator has a non-zero value at x= -2, there is a vertical asymptote there. Now, what value of p will guarentee that x= -2 is NOT a vertical asymptote?

11. Mar 1, 2008

### Charlie_russo

Say you had a number set 1-10. X can equal any of these numbers other than 5, how would you show that?

Hint: <,>

12. Mar 1, 2008

### HallsofIvy

Since the only possible vertical asymptote would be at x= -2, I have no idea what that has to do with this problem!

13. Mar 1, 2008

### symbolipoint

You want p to have a value? If p were to contain a factor of x^2 or higher power, then g(x) would have a slant asymtote; but that seems not exactly to be what you ask.

HallsofIvy - help us here maybe, since we seem to not be so advanced. What are we missing?

14. Mar 1, 2008

### Dick

p=(-6)?

15. Mar 1, 2008

### HallsofIvy

Not quite, Dick, you've lost track of sign somewhere.

At x= -2, the denominator, $x^3+ 2$ is equal to 0. Obviously, as I said above, as long as the numerator is not also 0, the value of the function goes to infinity as we approach x= -2: a vertical asymptote. The only way not to have a vertical asymptote is to make sure the numerator also goes to 0 at x= -2: that 3x+ p= 3(-2)+ p= -6+ p= 0 or p= 6. If p= 6, then
$$\frac{3x+ 6}{x^3+ 8}= \frac{3(x+ 2)}{(x+ 2)(x^2- 2x+ 4)}= \frac{3}{x^2- 2x+ 4}$$
Rather than a vertical asymptote, we have a curve with a "hole" at (-3, 3/7).

Last edited by a moderator: Mar 2, 2008