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

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

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- Thread starter teffy3001
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g(x) = (3x + p) / (x^3 + 8)

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

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HallsofIvy

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

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rock.freak667

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

and hence make the function tend to infinity?

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Dick

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

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symbolipoint

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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.

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Dick

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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.

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HallsofIvy

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

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Hint: <,>

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HallsofIvy

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

Hint: <,>

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symbolipoint

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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?

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Dick

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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

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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).

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