Question

Find the slope of the tangent line to the curve
3 x^2 + 6 xy + 3 y^3 = -1296
at the point ( -4,-8 ).

I came up with 24/184 and it is saying I am wrong.

Answer 1

6x + 6*1y + 6x*y' + 3*3y^2*y' = 0

Answer 2

now plug in x= -4, y = -8
6(-4) + 6(-8) + 6(-4)y' + 3*3(-8)^2*y' = 0

Answer 3

-72 -24y' +576y' = 0

Answer 4

552y' = 72
y' = 72/(552)
y' = 3/23

Answer 5

so you need to reduce to lowest terms, thats all

Answer 6

Oh wow, just needed to reduce? Thanks! Got it right.