Find equations of both lines that are tangent to the curve y

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

12x-y+10=0 -y=-12x-10 y=12x+10 slope = 12 Any line parallel to it will have slope 12. y=x^3 y\' = 3x^2 3x^2=12 x^2=4 x=+/-2 when x=-2, y=-8 when x=2, y=8 You need two points to determine the equation. Let (x1,y1) be the points. y-y1 = 12 (x-x1) is the equation. y-2 = 12 (x-8) y+2=12(x+8)