Find an equation of the line that is tangent to the graph of f(x)=x^3 and parallel to the line: 12x-y+5=0
Rearrange the second equation into y=mx+c to find the gradient
$\displaystyle y = 12x+5$ therefore the gradient is 12
If the tangent [ie: f'(x)] is parallel to this line it must be equal.
Find f'(x) and set it equal to 12 then get x.
Spoiler:
Put your value(s) of x into f(x) to get a co-ordinate.
Spoiler:
You can now use the equation of a straight line $\displaystyle y-y_1 = m(x-x_1)$
To my knowledge I am not sure if you can get an explicit solution