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**These are my steps (Please do the problem, but also tell me where my mistake was):**

Step 1: Deriv of curve

y'=3x^2

Step 2: Slope of Parallel line

m=12

Step 3: Plug 12 into y' for slope

12=3x^2

x= 2,-2

Step 4: Plug 2 into Original Equation to get points.

y(2)=1+8

y(2)=9

y(-2)=1-8

y(-2)=-7

Points= (2,9);(-2,-7)

Step 5: Plug points into y=mx+B equation with m=12 to get the equation of the lines tangent to y=1+x^3 but parallel to 12x-y

9=12(2)-1+b

9=24-1+b

b=-14

-7=12(-2)+b

b=-31

Step 6: Write equations of the lines because I found b.

y=12x-14 and y=12x-31