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

The tangent of f(x)=-3x^3-2x^2-1,\at-1 is y=(-9a_{0}^2-4a_{0})x+6a_{0}^3+2a_{0}^2-1

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

tangent of

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

y = (- 9 a_{0}^{2} - 4 a_{0}) x + 6 a_{0}^{3} + 2 a_{0}^{2} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 3 a_{0}^{3} - 2 a_{0}^{2} - 1)

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 = a_{0} : m = - 9 a_{0}^{2} - 4 a_{0}

Find the line with slope m=

- 9 a_{0}^{2} - 4 a_{0}

and passing through

(a_{0}, - 3 a_{0}^{3} - 2 a_{0}^{2} - 1) : y = (- 9 a_{0}^{2} - 4 a_{0}) x + 6 a_{0}^{3} + 2 a_{0}^{2} - 1

y = (- 9 a_{0}^{2} - 4 a_{0}) x + 6 a_{0}^{3} + 2 a_{0}^{2} - 1

\frac{s + 3}{s + 1 ( s + 2 )}

\int - 2 sin^{2} (t) d t

\frac{\partial}{\partial x} (4374 x^{\frac{1}{3}} + 2 y)

x \to 1 lim (\frac{e ^{x} - 1}{x})

\frac{\partial}{\partial z} (x ln (y + z))

What is the tangent of f(x)=-3x^3-2x^2-1,\at-1 ?
The tangent of f(x)=-3x^3-2x^2-1,\at-1 is y=(-9a_{0}^2-4a_{0})x+6a_{0}^3+2a_{0}^2-1