What is the tangent of x^4-20x^2+64,\at x=2 ?

The tangent of x^4-20x^2+64,\at x=2 is y=-48x+96

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tangent of x^4-20x^2+64,\at x=2

tangent of

x^{4} - 20 x^{2} + 64, at x = 2

y = - 48 x + 96

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

y = x^{4} - 20 x^{2} + 64 : \frac{d y}{d x} = 4 x^{3} - 40 x

Compute the slope of

y = x^{4} - 20 x^{2} + 64

x = 2 : m = - 48

Find the line with slope m=

- 48

and passing through

(2, 0) : y = - 48 x + 96

y = - 48 x + 96

2 t y^{^{'}} + 4 y = 3

t (t^{2} - 1)^{2}

\int 1 - r^{2} \cdot r d r

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

\int_{0}^{10} 20 x e^{- 0.6 x} d x

What is the tangent of x^4-20x^2+64,\at x=2 ?
The tangent of x^4-20x^2+64,\at x=2 is y=-48x+96