Can you help me?

Question

find the equation of the tangent line to the curve y = f (x) at the given point. y=f(x)=1-x³, at (-2,9)

Answer 1

y= f(x) = 1-x^3

f'(x) = -3x^2

f'(-2) = -3(-2)^2 = -12 = slope of a tangent line at the point (-2,9)

y -9 = -12(x +2) is the equation of the tangent line at the point (-2,9) in point slope form

y =-12x -24 +9

y = -12x -15 is the equation in slope intercept form

12x + y =-15 is the equation in standard form

12x +y +15 is the equation in general form