Heirio
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Tried looking it up, but nothing really answers my question.

How do you know which side the line is on when sketching the curve?
For example, with -1/x, the curves are in the top left and the bottom right quadrants. With 1/x, however, they're in the top right and the bottom left.

So is the fact that it's now negative affecting it?

Well, with (x-2)/x, they're in the bottom right and the top left. But with (x+2)/x, they're in the top right and the bottom left.

The idea I'm getting from all this is that the quadrants the curves generally lie in (excluding the location of the x and y asymptotes, I think I've got those nailed) is determined by the constant, the value without any x's. Namely whether it's positive or negative. I've tested it with different values and they only seem to make it more or less slopy, for the most part.

So, is that assessment correct? Does the sign of the numerator constant (the one without any x's, if that terminology is confusing) determine the quadrants the curve is in (irrespective of x and y asymptotes)?
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ghostwalker
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(Original post by Heirio)
Tried looking it up, but nothing really answers my question.

How do you know which side the line is on when sketching the curve?
For example, with -1/x, the curves are in the top left and the bottom right quadrants. With 1/x, however, they're in the top right and the bottom left.

So is the fact that it's now negative affecting it?

Well, with (x-2)/x, they're in the bottom right and the top left. But with (x+2)/x, they're in the top right and the bottom left.

The idea I'm getting from all this is that the quadrants the curves generally lie in (excluding the location of the x and y asymptotes, I think I've got those nailed) is determined by the constant, the value without any x's. Namely whether it's positive or negative. I've tested it with different values and they only seem to make it more or less slopy, for the most part.

So, is that assessment correct? Does the sign of the numerator constant (the one without any x's, if that terminology is confusing) determine the quadrants the curve is in (irrespective of x and y asymptotes)?
In the case of (x+a)/x, yes, the sign of the constant determines which quadrants.

If we divide out, we get 1 + a/x. The 1 only shifts the curves "up" or "down".

So you essentially have a/x determining the quadrants.
Postive, then top right and bottom left.
Negative, then top left, and bottom right.

Also, a/x will look roughtly the same as 1/x, if a is positive - a is just a scaling factor really.

And it will look the same as -1/x if a is negative.

If you have any doubts, try plotting a point or two. Choose an easy one x=1, or -1, or whatever's convenient.
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Heirio
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What about other degrees? Like if you have a quadratic on the top?
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ghostwalker
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(Original post by Heirio)
What about other degrees? Like if you have a quadratic on the top?
Plotting some points will tell you which area, above or below, you're dealing with.
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