What is the tangent of y=x^3,\at x=-9 ?

The tangent of y=x^3,\at x=-9 is y=243x+1458

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tangent of y=x^3,\at x=-9

tangent of

y = x^{3}, at x = - 9

y = 243 x + 1458

Solution steps

Find the tangent point:

(- 9, - 729)

Find the slope of

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

Compute the slope of

y = x^{3}

x = - 9 : m = 243

Find the line with slope m=

243

and passing through

(- 9, - 729) : y = 243 x + 1458

y = 243 x + 1458

(1 + sin (x))^{- 4}

\int (\frac{1}{x} + \frac{1}{3 - x}) d x

x \to 0 lim (sin (5 x) csc (3 x))

0.02 (1 + 3 cos (t))

\frac{d y}{d t} = 25 - 10 t - 2 y

What is the tangent of y=x^3,\at x=-9 ?
The tangent of y=x^3,\at x=-9 is y=243x+1458