Question

find points where the tangent line is horizontal
y=(-1/3x^3)+(6x^2)-11x-50

Answer 1

y'=0

Answer 2

i need the coordinates where the tangent line is zero

Answer 3

srry kind of was missing parenthesis http://www.wolframalpha.com/input/?i=%28%28-1%2F3*x^3%29%2B%286x^2%29-11x-50%29%29%27%3D0

Answer 4

Do you mean:\[-(1/3)x^3 + 6x^2 - 11x -50\]?

If so take the first derivative and set equal to 0\[0 = -(1/3)*3x^2+6*2x-11\]

Answer 5

kk how did you get both coords?

Answer 6

the x coords are given by solving y'=0, ther are 1 and 11, the y ccords are given by puting those values in the original function

Answer 7

hmm i got -1 and 11

Answer 8

kk thx guys