Points on the graph of:

y = (1/3)x^3 + x^2 - x

where the tangen has slope 1, are the points where dy/dx=1.

So you need to differentiate (1/3)x^3 + x^2 - x with respect to x, and then

set the derivative equal to zero and solve the resulting equation. This will

give the x values where the slope is 1, the y values are obtained by

substituting these x's back into y = (1/3)x^3 + x^2 - x.

RonL