Calculus: The tangent line intersects a curve at two points. Find the other point. The line tangent to $y = -x^3 + 2x + 1$ when $x = 1$ intersects the curve in another point.  Find the coordinates of the other point.
This was never taught in class, and I have a test on this tomorrow.  This question came off of my test review worksheet, and I don't understand how to solve it.  The answers are on the back, and for this one it says the answer is (-2,5), but I don't understand how to get that.
I did the derivative and substituted 1 for x to get the slope of the line:
$y = -x^3 + 2x + 1$
$y' = -3x^2 + 2$
$y' = -3 + 2$
$y' = -1$
I don't know where to go from here.

If we let $f(x)=-x^3+2x+1$, then we know the derivative at $x=1$, namely $f'(1)=-1$.  Now we know the slope of the tangent line $t(x)$.  In particular, this implies $t(x)=-x+c$ for some constant $c \in \mathbb{R}$.
We can also compute a point on the line, namely $(1,f(1))$.  This gives the value of $t(1)=f(1)$, which can be used to solve for $c$.
Then find the other point where $f(x)$ and $t(x)$ intersect, i.e., where $f(x)-t(x)=0$.
