- #1

- 23

- 0

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

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter teffy3001
- Start date

- #1

- 23

- 0

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

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

- #2

- 980

- 2

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

- #4

- 23

- 0

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

- #5

rock.freak667

Homework Helper

- 6,230

- 31

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

and hence make the function tend to infinity?

- #6

- 23

- 0

- #7

Dick

Science Advisor

Homework Helper

- 26,260

- 619

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

- #8

symbolipoint

Homework Helper

Education Advisor

Gold Member

- 6,234

- 1,219

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

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

Dick

Science Advisor

Homework Helper

- 26,260

- 619

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.

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

- #10

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

Yes, at x= -2, the denominator goes to 0. As long as the numerator has a

- #11

- 16

- 0

Hint: <,>

- #12

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

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

Hint: <,>

- #13

symbolipoint

Homework Helper

Education Advisor

Gold Member

- 6,234

- 1,219

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

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

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

Dick

Science Advisor

Homework Helper

- 26,260

- 619

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?

p=(-6)?

- #15

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

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

At x= -2, the denominator, [itex]x^3+ 2[/itex] 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

[tex]\frac{3x+ 6}{x^3+ 8}= \frac{3(x+ 2)}{(x+ 2)(x^2- 2x+ 4)}= \frac{3}{x^2- 2x+ 4}[/tex]

Rather than a vertical asymptote, we have a curve with a "hole" at (-3, 3/7).

At x= -2, the denominator, [itex]x^3+ 2[/itex] 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

[tex]\frac{3x+ 6}{x^3+ 8}= \frac{3(x+ 2)}{(x+ 2)(x^2- 2x+ 4)}= \frac{3}{x^2- 2x+ 4}[/tex]

Rather than a vertical asymptote, we have a curve with a "hole" at (-3, 3/7).

Last edited by a moderator:

Share:

- Replies
- 3

- Views
- 9K

- Replies
- 3

- Views
- 816