Find the slope of the tangent line to the graph of f(x)=(-x^2+4)^3 at x=1

Answer 1

f(x) = (-x^2+4)^3

f(1) = (-1^2+4)^3 = 3^3 = 27

f'(x) =3(-x^2+4)(-2x)

f'(1) =3(3)(-6)= -54

y-27 = -54(x-1) = -54x +54

y = -54x +65+27

y =-54x + 81 is the equation in slope intercept form, with slope=-54 and y intercept = 81

Answer 2

f(1)=27

f'(x)=3(-x²+2)²(-2x)

f'(1)=-54

The equation of the tangent is -54=(y-27)/(x-1)

You can "massage" the form into any needed form.