What is the tangent of f(x)=-3x^3-2x^2-1,\at x=-1 ?

The tangent of f(x)=-3x^3-2x^2-1,\at x=-1 is y=-5x-5

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tangent of f(x)=-3x^3-2x^2-1,\at x=-1

tangent of

f (x) = - 3 x^{3} - 2 x^{2} - 1, at x = - 1

y = - 5 x - 5

Solution steps

Find the tangent point:

(- 1, 0)

Find the slope of

f (x) = - 3 x^{3} - 2 x^{2} - 1 : \frac{df ( x )}{d x} = - 9 x^{2} - 4 x

Compute the slope of

f (x) = - 3 x^{3} - 2 x^{2} - 1

x = - 1 : m = - 5

Find the line with slope m=

- 5

and passing through

(- 1, 0) : y = - 5 x - 5

y = - 5 x - 5

\frac{\partial}{\partial x} (\frac{x z}{x ^{2} + y ^{2}})

\int cos (3 x) (2 + sin (3 x))^{4} d x

\frac{d}{d a} (sin (a) + sin (2 a))

\frac{d}{d x} (\frac{23}{x ^{4}})

\int \frac{1}{tan ( 5 θ )} d θ

What is the tangent of f(x)=-3x^3-2x^2-1,\at x=-1 ?
The tangent of f(x)=-3x^3-2x^2-1,\at x=-1 is y=-5x-5