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
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?
2 answers
The answer is 0