<s>[INST] Calculus - Derivatives and Tangent Lines I am trying to find the equation of the lines (there's 2) tangent to the graph of $f(x) = x^3$ and parallel to the line $12x − y + 1 = 0$. 
So far obviously the slope of the line must be $12$.
The equation of the line tangent to $f(x) = x^3$ is the derivative and must be equal to the slope of the intersecting line, $12$. The derivative of $x^3$ is equal to $2x^2$, so $2x^2 = 12$, and $x=$ $\pm\sqrt{6}$, so $y$ must equal to $\pm6\sqrt{6}$ and thus the equation of the line is $y=12x-6\sqrt{6}$ or $y=12x+6\sqrt{6}$. Is this correct? Thanks.
EDIT
I think I made a mistake it should be  $y=12x-5\sqrt{6}$ or $y=12x+5\sqrt{6}$
EDIT again 
Thanks to everyone who helped out!
The derivative is equal to $3x^2$ not $2x^2$, so $x =\pm 2$ and so $y =\pm 8$ so plugging back into the equation gives $y=12x-16$ or $y=12x+16$, right?
 [/INST] Yes, the slope of our line(s) needs to be $12$ to ensure they are parallel to the line
$\,12x − y + 1 = 0.$
Next, finding the derivative of $y = x^3$ gives us $\,y' = 3x^2$. 
To find the points on the curve at which the tangent lines at those points have slope $12$, we solve $$3x^2 = 12 \iff x^2 = 4 \iff x =\pm 2.$$
Following your edits: Yes, you are correct. 
There are two lines tangent to the curve $y = x^3$ with slope $12$. One point at which a line with slope $12$ is tangent is given by $(-2, -8)$ and the other point at which a line with slope $12$ is tangent to the curve is given by $(2, 8)$.
So you have slope $m = 12$, and a point $(x_0, y_0)$ for each of two such tangent lines.
$$(y - y_0) = m(x - x_0)$$
Hence, one line is given by $$y + 8 = 12(x + 2) \iff y = 12 x + 16$$
The other line is given by $$y - 8 = 12(x -2) \iff y = 12x - 16$$
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