What is the tangent of 8x^2-2,\at x=-4 ?

The tangent of 8x^2-2,\at x=-4 is y=-64x-130

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tangent of 8x^2-2,\at x=-4

tangent of

8 x^{2} - 2, at x = - 4

y = - 64 x - 130

Solution steps

Find the tangent point:

(- 4, 126)

Find the slope of

y = 8 x^{2} - 2 : \frac{d y}{d x} = 16 x

Compute the slope of

y = 8 x^{2} - 2

x = - 4 : m = - 64

Find the line with slope m=

- 64

and passing through

(- 4, 126) : y = - 64 x - 130

y = - 64 x - 130

x^{4} + 2 x^{2} + 3, at x = - 2

f (x) = 4 e^{2 x^{3}}

\int_{- \infty}^{0} \frac{1}{7 - 4 x} d x

\int x^{2} e^{- 5 x} d x

\frac{\partial}{\partial x} (\frac{y + 1}{x})

What is the tangent of 8x^2-2,\at x=-4 ?
The tangent of 8x^2-2,\at x=-4 is y=-64x-130