Find the coordinates of the point on the curve $f(x)=3x^2-4x$ where the tangent is parallel to the line $y=8x$.
The derivative of the function $f$ tells you the slope of the tangent to $f$ at the point $(x, f(x))$. Since the tangent has to be parallel to the line $y = 8x$, it is clear that the slope must be $8$. Therefore, $$f'(x) = 8 \implies 6x - 4 = 8 \implies x = 2$$
The point is then $(2, 4)$.