Equation of a line that has to be tangent

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How would find the equation of a line that has to be tangent to a curve and parallel to another line (i know slope has to be equal)

(you are given the equation of the line and the curve)
x^3
3x-y-6
 
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Ok so you know the gradients have to be the same, so wherever the gradient is 3 on the curve (finding where this happens will be done by using the derivative) will be used to find the equation of the new line.
Taking the derivative of the curve will give you y'=3x^2

and since this has to be equal to the gradient of the line which is 3, just solve 3x^2=3.
Notice you'll have 2 answers, and that is how it should be! :smile:

And use the point-gradient form for the equation of a line:
y-y_1=m(x-x_1)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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