what is the equation for the tangent to the curve 6y=7-x^3 at (1,1)?
6 dy/dx = -3x^2
at (1,1)
6 dy/dx = -3
dy/dx = -3/6 = -1/2
slope = -1/2, point is (1,1)
find the equation using the method that you learned.
at (1,1)
6 dy/dx = -3
dy/dx = -3/6 = -1/2
slope = -1/2, point is (1,1)
find the equation using the method that you learned.