
Intersect points
I was given this question: Find the tangent through point (1,1) of y^2 = x^3 2x + 2. After some time i found out that this is (x3)/2. But the b) part of the question was: the curves intersect at another point, find this? And i couldnt because setting both equations equal to each other gives me 1,1 again..? Thank you!

Re: Intersect points
If $\displaystyle y^2= x^3 2x+ 2$, then $\displaystyle 2yy'= 3x^2 2$ so that at (1, 1), 2y'= 1 and y'= 1/2. No, (x 3)/2 is incorrect because that has slope 1/2, not 1/2.
Yes, one solution is (1, 1) again but that will be a quadratic equation so there may be another solution.

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Re: Intersect points

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Re: Intersect points
Your first question makes perfect sense. But the second is completely mysterious to me. What curves (plural)? The answer above is one interpretation  namely, find a second point where the tangent line intersects the curve. By the way, there are several good free or inexpensive graphing programs available, Geogebra and Graph, for example. A little time spent "checking your work" with such a program is far faster than online help. I rapidly created the following:
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Re: Intersect points
The second curve is the straight line.