What is the tangent of x^3,\at x=-2 ?

The tangent of x^3,\at x=-2 is y=12x+16

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tangent of x^3,\at x=-2

tangent of

x^{3}, at x = - 2

y = 12 x + 16

Solution steps

Find the tangent point:

(- 2, - 8)

Find the slope of

y = x^{3} : \frac{d y}{d x} = 3 x^{2}

Compute the slope of

y = x^{3}

x = - 2 : m = 12

Find the line with slope m=

12

and passing through

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

y = 12 x + 16

\frac{s - 7}{s + 5}

\int tan (7 x se) c^{2} 7 x d x

\int e^{- 2 x} cos (3 x) d x

x \to 3 lim (x^{2} + 16)

\frac{\partial}{\partial x} (25 - 5 x^{2} - y^{2})