Question

Write the equation of the tangent of the line to the curve y=x^3+3x-8 at (2,6)

Answer 1

y' = 3x^2 + 3

m = 3(2)^2 + 3 = 15

y - 6 = 15(x - 2)
y = 15x -24

Answer 2

ok. Thanks so much. I was just making sure I was correct. So you took the derivative and then plugged in the 2 in order to get slope. After you used the point-slope formula to get the equation of the tangent of the line :)