What is the tangent of f(x)=-4x^2-4x+4,\at x=1 ?

The tangent of f(x)=-4x^2-4x+4,\at x=1 is y=-12x+8

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tangent of f(x)=-4x^2-4x+4,\at x=1

tangent of

f (x) = - 4 x^{2} - 4 x + 4, at x = 1

y = - 12 x + 8

Solution steps

Find the tangent point:

(1, - 4)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 12

Find the line with slope m=

- 12

and passing through

(1, - 4) : y = - 12 x + 8

y = - 12 x + 8

- \frac{12}{( 4 x + 1 ) ^{2}}

\int \frac{1}{( e ^{y} + 1 ) ^{2} e ^{- y}} d y

\frac{d y}{d x} = \frac{4 cos ( 2 x )}{3 - sin ( 2 x )}

\frac{10}{s ( s ^{2} + 8 s + 25 )}

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

What is the tangent of f(x)=-4x^2-4x+4,\at x=1 ?
The tangent of f(x)=-4x^2-4x+4,\at x=1 is y=-12x+8