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 \frac{( 9 sin ( x ) + 4 sec ( x ) )}{tan ( x )} d x

\frac{2}{x}

1 (t - 4)

x \to 0 + lim (sin^{x} (x))

x \to 1 lim (2 x^{2} + x - 1)

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