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