What is the tangent of f(x)=-x^3,(-2,8) ?

The tangent of f(x)=-x^3,(-2,8) is y=-12x-16

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tangent of f(x)=-x^3,(-2,8)

tangent of

f (x) = - x^{3}, (- 2, 8)

y = - 12 x - 16

Solution steps

Find the slope of

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

Compute the slope of

f (x) = - x^{3}

(- 2, 8) : m = - 12

Find the line with slope m=

- 12

and passing through

(- 2, 8) : y = - 12 x - 16

y = - 12 x - 16

\int cos (x^{2}) 2 x d x

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

\int \frac{e ^{5 x}}{e ^{5 x} + 6} d x

- 2 x y - 4 x^{4} = - 2 + y

\frac{\partial}{\partial x} (12 x y - x^{3} - y^{3} - 7)

What is the tangent of f(x)=-x^3,(-2,8) ?
The tangent of f(x)=-x^3,(-2,8) is y=-12x-16