Intersect points

• Dec 31st 2012, 09:10 AM
jones123
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 (x-3)/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!
• Dec 31st 2012, 10:26 AM
HallsofIvy
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.
• Jan 1st 2013, 03:57 AM
ibdutt
Re: Intersect points
• Jan 1st 2013, 09:02 AM
johng
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:
Attachment 26436
• Jan 1st 2013, 07:44 PM
ibdutt
Re: Intersect points
The second curve is the straight line.