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

The tangent of f(x)=-2x^2+4x-4,\at x=3 is y=-8x+14

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tangent of f(x)=-2x^2+4x-4,\at x=3

tangent of

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

y = - 8 x + 14

Solution steps

Find the tangent point:

(3, - 10)

Find the slope of

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

Compute the slope of

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

x = 3 : m = - 8

Find the line with slope m=

- 8

and passing through

(3, - 10) : y = - 8 x + 14

y = - 8 x + 14

\frac{\partial}{\partial x} (x + 3 y)

(lo g_{e} (x))^{^{'}}

(- 3, 5), (3, 5)

y^{^{'}} = 2 x + 3

64 y^{^{''}} - 48 y^{^{'}} + 34 y = 0

What is the tangent of f(x)=-2x^2+4x-4,\at x=3 ?
The tangent of f(x)=-2x^2+4x-4,\at x=3 is y=-8x+14