Asked by Arnold
                The slope of the tangent line to the graph y= x^3/3-x at the point (1, -2/3)
So I plugged this into y= and I found the (1,-2/3) point but I don't know what to do next, can you help?
            
        So I plugged this into y= and I found the (1,-2/3) point but I don't know what to do next, can you help?
Answers
                    Answered by
            Steve
            
    the tangent line has the same slope as the curve at the point of tangency. The slope at any x is
y' = x^2-1
So, y'(1) = 0
Now you have a point and a slope, so the line is
y + 2/3 = 0(x-1)
or,
y = -2/3
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D+x^3%2F3-x%2C+y%3D-2%2F3
    
y' = x^2-1
So, y'(1) = 0
Now you have a point and a slope, so the line is
y + 2/3 = 0(x-1)
or,
y = -2/3
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D+x^3%2F3-x%2C+y%3D-2%2F3
                    Answered by
            Anonymous
            
    The answer is 0
    
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