What is the tangent of y=-4x^2,(-2,-16) ?

The tangent of y=-4x^2,(-2,-16) is y=16x+16

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tangent of y=-4x^2,(-2,-16)

tangent of

y = - 4 x^{2}, (- 2, - 16)

y = 16 x + 16

Solution steps

Find the slope of

y = - 4 x^{2} : \frac{d y}{d x} = - 8 x

Compute the slope of

y = - 4 x^{2}

(- 2, - 16) : m = 16

Find the line with slope m=

16

and passing through

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

y = 16 x + 16

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\frac{\partial}{\partial x} ((3 x - 2 x^{2}) 3)

\int (\frac{2 x}{x ^{2} + 1}) d x

What is the tangent of y=-4x^2,(-2,-16) ?
The tangent of y=-4x^2,(-2,-16) is y=16x+16