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Find a vector U that is tangent to the graph of y=x^2-x+3 at the point (2,5)
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slope is 5 vector is |dw:1327542825433:dw|
The answer is actually <1, 3> and I am asking for an explanation why
oh...sry, that's right
Using derivatives, if you get the first derivative of the quadratic above, we get slope=2x-1 when we plug 2 in, we get slope=3 put this in vector form, we get (1,3)
added 1 by mistake on the first try :(
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ah, that simple. Thanks!
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