I was given the function $y = x^3-x$ and told to find all tangent lines that pass through $(-2,2)$. Not sure what steps to take past finding the derivative.
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$\begingroup$We find the equation of the tangent line to $f(x) = x^3-x$ at the point $(k, k^3-k)$.
First, the derivative gives the slope $$f'(x) = 3x^2 -1$$
So we have a line with slope $3k^2-1$ that goes through the point $(k,k^3-k)$. What is its equation? We use the point-slope form
$$y-(k^3-k) = (3k^2-1)(x-k)$$
$$y-k^3 +k = 3k^2x-x-3k^3+k$$
$$y+2k^3 = 3k^2x-x$$
We want to know which values of $k$ make this line pass through $(x,y)=(-2,2)$, so we plug in those values:
$$2+2k^3=3k^2(-2)-(-2)$$
$$2k^3 = -6k^2$$
$$k^3 = -3k^2$$
$$k=0 \,\,\,\,\text{ or }\,\,\, k=-3$$
This means that there are two tangent lines that work, one at $(0,0)$, and the other at $(-3,-24)$. You should be able to find the equations of these lines easily if necessary.
$\endgroup$ 2 $\begingroup$For any real $x_0$, you can compute the equation of the tangent of the function that passes thru the point $(x_0,f(x_0))$. These equations are of the form $y = a(x_0)x + b(x_0)$, where $a$ and $b$ are functions that you have to compute. Now, replace $x = -2$ and $y=2$ gives you an equation on $x_0$; the solutions are exactlly the $x_0$ such that the tangent line thru $x_0$ passes thru $(-2,2)$.
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