Find all line equations that are tangent to $x^3 - x$ and pass through $(-2,2)$ So I have the equation: 
$f(x) = x^3 - x$
So we know that the slope of the curve for some $x$ is given by: 
$f'(x) = 3x^2 - 1$
And need to find equations of lines that are tangent to that curve, and also pass through the point $(-2,2)$.
I have seen a similar question involving a circle equation, but couldn't get my head around it.
 A: Given a point on the curve, it has the form $(a,a^3-a)$.  And the slope of the tangent at that point is $3a^2-1$.  Then, using point-slope form for a line, the tangent line has the form
$$
y=(3a^2-1)(x-a)+a^3-a.
$$
Expanding and simplifying, this becomes
$$
y=(3a^2-1)x-2a^3.
$$
If $(-2,2)$ is on this line, we must have
$$
2=(3a^2-1)(-2)-2a^3.
$$
A: the eqution of line is 
$$\frac{y-y_0}{x-x_0}=slope$$
$$\frac{x^3-x-2}{x+2}=3x^2-1$$
$$3x^3-x+6x^2-2=x^3-x-2$$
$$2x^3+6x^2=0$$
$$2x^2(x+3)=0$$
if $$x=0$$
or $$x=-3$$
that means there are two lines tangent to the $x^3-x$

A: There are two approaches to this: you can find all the tangents that pass through $(-2,2)$, or find all the lines through $(-2,2)$. Both should end up with the same equations.
All lines that pass through $(-2,2)$ are of the form
$$ y-2 = m(x+2). $$
These intersect the curve $y=x^3-x$ when
$$ x^3-x-2 = m(x+2). $$
For an intersection to be a tangent, the gradient has to match as well, i.e.
$$ 2x^2-1 = m, $$
which has solutions
$$ x = \pm\sqrt{\tfrac{1}{2}(m+1)}. $$
Perhaps more sensible is to eliminate $m$ and find
$$ 0 = x^3-x-2 - (2x^2-1)(x+2) = -x^3-4x^2 = -x^2(x+4), $$
so $x=0$ or $x=-4$, and then you know that $m=2x^2-1$, which you can use to find the gradients.

Going the other way, the equation of the general tangent is
$$ y'-(x^3-x) = (2x^2-1)(x'-x), $$
because this is the line that passes through $(x,x^3-x)$ with gradient $2x^1-1$. If we now demand that this line pass through $(-2,2)$, i.e. set $y'=2$, $x'=-2$, we get
$$ 2-(x^3-3) = (2x^2-1)(-2-x), $$
which is equivalent to the equation we got doing it the other way.
