Find the equation of tangents to the curve #y=x^3# which are parallel to the line #12x-y-3=0#?
1 Answer
Jun 15, 2017
Explanation:
The line
So we need to draw a tangent to
As slope at a point is given by first derivative of the function i.e.
we should have
Hence wewill have such a tangent at two points,
one at
and tangents will be
graph{(y-x^3)(12x-y+16)(12x-y-16)(12x-y-3)=0 [-5, 5, -15, 15]}