What is the tangent of f(x)=2x^4-4x^2+6,\at x=-2 ?

The tangent of f(x)=2x^4-4x^2+6,\at x=-2 is y=-48x-74

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tangent of f(x)=2x^4-4x^2+6,\at x=-2

tangent of

f (x) = 2 x^{4} - 4 x^{2} + 6, at x = - 2

y = - 48 x - 74

Solution steps

Find the tangent point:

(- 2, 22)

Find the slope of

f (x) = 2 x^{4} - 4 x^{2} + 6 : \frac{df ( x )}{d x} = 8 x^{3} - 8 x

Compute the slope of

f (x) = 2 x^{4} - 4 x^{2} + 6

x = - 2 : m = - 48

Find the line with slope m=

- 48

and passing through

(- 2, 22) : y = - 48 x - 74

y = - 48 x - 74

\frac{\partial}{\partial z} (tan (1 + 2 x^{2} y^{3} z^{2}))

\int_{- π}^{π} e^{- in x} d x

f (x) = \frac{4}{x}

\frac{d}{d x} (\frac{3 x ( 4 - x )}{5 x ^{2}})

x \to 0 lim (\frac{sin ( 3 x )}{x} + 2)

What is the tangent of f(x)=2x^4-4x^2+6,\at x=-2 ?
The tangent of f(x)=2x^4-4x^2+6,\at x=-2 is y=-48x-74