Question

PLEASE help me find equations of both lines that are tangent to the curve y = 1 + x^3 and are parallel to the line 12x-y=1

Answer 1

12x-y=1
y= 12x-1 so the line has a slope of +12

find dy/dx of the curve, and solve for the 2 x values where dy/dx =12
Is that enough of a hint?

Answer 2

well so (dy/dx)(1+x^3) = 3x^2, so what where x = 2 ... ? so what's the answer?

Answer 3

This video might help you a lot
http://www.khanacademy.org/math/calculus/v/calculus--derivatives-1--new-hd-version

meanwhile, you found
dy/dx = 3x^2

dy/dx represents the change in y over change in x (slope!) which changes at every x. You want the x (actually 2 x's) where dy/dx = +12
find the x, then find the y (use y= 1+x^3)
now you have a point (x,y) and a slope, so you can find the equation of a line through that point with slope 12

Answer 4

got it thank you