What is the tangent of y=-2x^3,(-2,16) ?

The tangent of y=-2x^3,(-2,16) is y=-24x-32

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tangent of y=-2x^3,(-2,16)

tangent of

y = - 2 x^{3}, (- 2, 16)

y = - 24 x - 32

Solution steps

Find the slope of

y = - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2}

Compute the slope of

y = - 2 x^{3}

(- 2, 16) : m = - 24

Find the line with slope m=

- 24

and passing through

(- 2, 16) : y = - 24 x - 32

y = - 24 x - 32

x \to 0 lim ((1 - 2 x)^{\frac{4}{x}})

\int \frac{2 y}{3} d y

\int \frac{( y - 4 )}{y ^{2} - 8 y + 10} d y

f (x) = \frac{5}{x}, at x = a

f (x) = x^{4} + 4 x^{2} - x, at x = 1

What is the tangent of y=-2x^3,(-2,16) ?
The tangent of y=-2x^3,(-2,16) is y=-24x-32