- #1

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I got (2,1) as a point. Are there more than one, or is that the only one?

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- Thread starter BrownianMan
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- #1

- 134

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I got (2,1) as a point. Are there more than one, or is that the only one?

- #2

rock.freak667

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So for each 'x' you pick, there will be a corresponding value of 'y' from that relation. Yes (2,1) is a point, so is (-2,1)

- #3

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- #4

Mark44

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For the curve y^2 + x^3 = 9 (it's not a line, as you stated), y' = -3x^2/(2y).

The slope of the specified tangent line is -6.

Equating those values gives x^2 = 4y ==> x = +/-2sqrt(y)

Now substitute for x in the equation of the tangent line y = -6x + 13. This gives you two values of y, one of which is y = 1. The other is extraneous.

Now substitute this y value in the equation of your curve y^2 + x^3 = 9, and you get only one x-value.

(-2, 1) is not a point on your curve.

- #5

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So the only point is (2,1) as I stated in my first post, right?

- #6

Mark44

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- #7

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Thank you so much!

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