For the curve y=2x^3, the equation of the tangent that crosses a point (x,y) is y=8x + b. Determine the value of b
the slope of any tangent is y' = 6x^2
So, at the point (h,k), the tangent line has equation
y-k = 6h^2(x-h)
But k = 2h^3, so
y = 6h^2(x-h) + 2h^3
y = 6h^2 x - 4h^3
6h^2 = 8, so h=√(4/3)
so, b = -4(4/3)^(3/2)
y = 8x - 4(4/3)^(3/2)
Note that our point of tangency is at (2/√3, 16/3√3) ≈ (1.15,3.08)
See the graphs at
www.wolframalpha.com/input/?i=plot+y%3D2x%5E3%2C+y+%3D+8x+-+4%284%2F3%29%5E%283%2F2%29+for+x%3D0..2