What is the tangent of y=x^3,(-2,-8) ?

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

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tangent of y=x^3,(-2,-8)

tangent of

y = x^{3}, (- 2, - 8)

y = 12 x + 16

Solution steps

Find the slope of

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

Compute the slope of

y = 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

n = 1 \sum \infty \frac{( - 3 ) ^{n}}{n}

\frac{d}{d x} (x - x^{- 1})

y^{^{'}} - \frac{3}{x} y = \frac{y ^{4}}{x ^{2}}

\frac{d}{d s} (s i)

\frac{x ^{2} + 1}{x}