What is the tangent of f(x)=x^3-3x^2-9x+5,\at x=-12 ?

The tangent of f(x)=x^3-3x^2-9x+5,\at x=-12 is y=495x+3893

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tangent of f(x)=x^3-3x^2-9x+5,\at x=-12

tangent of

f (x) = x^{3} - 3 x^{2} - 9 x + 5, at x = - 12

y = 495 x + 3893

Solution steps

Find the tangent point:

(- 12, - 2047)

Find the slope of

f (x) = x^{3} - 3 x^{2} - 9 x + 5 : \frac{df ( x )}{d x} = 3 x^{2} - 6 x - 9

Compute the slope of

f (x) = x^{3} - 3 x^{2} - 9 x + 5

x = - 12 : m = 495

Find the line with slope m=

495

and passing through

(- 12, - 2047) : y = 495 x + 3893

y = 495 x + 3893

\int_{0}^{2} 0 d x

y^{^{'}} = (0.5 + sin (t)) (\frac{y}{5})

e^{- x}, x = 0

\int \frac{1}{x x - 4} d x

What is the tangent of f(x)=x^3-3x^2-9x+5,\at x=-12 ?
The tangent of f(x)=x^3-3x^2-9x+5,\at x=-12 is y=495x+3893