What is the tangent of f(x)=-4-3x^2,(2,-16) ?

The tangent of f(x)=-4-3x^2,(2,-16) is y=-12x+8

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tangent of f(x)=-4-3x^2,(2,-16)

tangent of

f (x) = - 4 - 3 x^{2}, (2, - 16)

y = - 12 x + 8

Solution steps

Find the slope of

f (x) = - 4 - 3 x^{2} : \frac{df ( x )}{d x} = - 6 x

Compute the slope of

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

(2, - 16) : m = - 12

Find the line with slope m=

- 12

and passing through

(2, - 16) : y = - 12 x + 8

y = - 12 x + 8

\int_{- 45}^{0} \frac{1}{4 - x} d x

n = 1 \sum \infty 8^{- n}

\frac{\partial}{\partial x} (1 + e^{- x y} cos (y))

\int 2 cos (4 x) cos (2 x) d x

\int 4 1 - x^{2} d x

What is the tangent of f(x)=-4-3x^2,(2,-16) ?
The tangent of f(x)=-4-3x^2,(2,-16) is y=-12x+8