I have two points on this graph. One is (1,0), the other is (0,∞). It looks like half a parabola or hyperbola.
The x value's range does not need to be just 1 to 0 - it may be drawn as 100 down to 0, but the y value stretches to infinity. As x approaches zero, y approaches infinity.
I know the function for the parabola is y = x², but I also know my graph only spans one unit - 0 to 1, and so it cannot be this. Am I wrong?
So I was wandering if anyone knows this function and could share it with me, and whether it has a name. Thanks
The x value's range does not need to be just 1 to 0 - it may be drawn as 100 down to 0, but the y value stretches to infinity. As x approaches zero, y approaches infinity.
I know the function for the parabola is y = x², but I also know my graph only spans one unit - 0 to 1, and so it cannot be this. Am I wrong?
So I was wandering if anyone knows this function and could share it with me, and whether it has a name. Thanks
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You don't really have enough information to answer this. ln(1/x) is a good guess, but there are infinitely many other functions that fit your description. For example, cot(pi*x/2).
Think of it this way: if you had only 2 "normal" points (not at infinity) on a curve, you couldn't tell if the curve was a straight line or a parabola or a circle or whatever. There are an infinite number of possibilities.
Think of it this way: if you had only 2 "normal" points (not at infinity) on a curve, you couldn't tell if the curve was a straight line or a parabola or a circle or whatever. There are an infinite number of possibilities.
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(0,∞) isn't really a point, because infinity is not a number.
But if you mean the graph passes through (1,0) and has a vertical asymptote at x = 0, you could do:
y = -tan((x+1)*pi/2)
The other Jeff's suggestion of ln(1/x) also works.
But if you mean the graph passes through (1,0) and has a vertical asymptote at x = 0, you could do:
y = -tan((x+1)*pi/2)
The other Jeff's suggestion of ln(1/x) also works.
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y = ln(1/x)