What is the tangent of f(x)=4x^3-12x^2,\at x=-2 ?

The tangent of f(x)=4x^3-12x^2,\at x=-2 is y=96x+112

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tangent of f(x)=4x^3-12x^2,\at x=-2

tangent of

f (x) = 4 x^{3} - 12 x^{2}, at x = - 2

y = 96 x + 112

Solution steps

Find the tangent point:

(- 2, - 80)

Find the slope of

f (x) = 4 x^{3} - 12 x^{2} : \frac{df ( x )}{d x} = 12 x^{2} - 24 x

Compute the slope of

f (x) = 4 x^{3} - 12 x^{2}

x = - 2 : m = 96

Find the line with slope m=

96

and passing through

(- 2, - 80) : y = 96 x + 112

y = 96 x + 112

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x \to 3 lim (5 + x)

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What is the tangent of f(x)=4x^3-12x^2,\at x=-2 ?
The tangent of f(x)=4x^3-12x^2,\at x=-2 is y=96x+112