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at what point on the curve x^2-y^2+x=2 is the tangent line vertical?
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can we take the derivative?
2x -2yy' +1 =0 -2x -1 ------ = y' y when y=0
forgot a couple details, but the result is the same
x^2 + x = 2 x^2 +x -2 =0 x = -4 and x= 2 the point (-4,0) (2,0)
doh....divide by 2
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(-2,0) and (1,0)
k, that is what i was going to tell you, thanks a lot
by all means, when im wrong let me know :)
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