What is the tangent of f(x)=4x^3+8x,(-1,-12) ?

The tangent of f(x)=4x^3+8x,(-1,-12) is y=20x+8

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tangent of f(x)=4x^3+8x,(-1,-12)

tangent of

f (x) = 4 x^{3} + 8 x, (- 1, - 12)

y = 20 x + 8

Solution steps

Find the slope of

f (x) = 4 x^{3} + 8 x : \frac{df ( x )}{d x} = 12 x^{2} + 8

Compute the slope of

f (x) = 4 x^{3} + 8 x

(- 1, - 12) : m = 20

Find the line with slope m=

20

and passing through

(- 1, - 12) : y = 20 x + 8

y = 20 x + 8

e^{(, at)}

(0, - 4), (3, - 7)

\frac{2 s ^{3} + 3 s ^{2} - 2 s}{s ^{3}}

m (t) = \frac{4 t + 11}{7 t ^{2} - 3 t + 5}

\frac{d}{d x} (\frac{x}{x + \frac{y}{x}})

What is the tangent of f(x)=4x^3+8x,(-1,-12) ?
The tangent of f(x)=4x^3+8x,(-1,-12) is y=20x+8