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

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

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

tangent of

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

y = 8

Solution steps

Find the tangent point:

(- 2, 8)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = 0

Find the line with slope m=

0

and passing through

(- 2, 8) : y = 8

y = 8

\frac{1 - 3 x + x ^{3}}{x ^{3}}

\frac{d}{d x} (sin (x y))

\frac{d}{d x} ((8 x - x^{2})^{3})

\int (1 - sin^{2} (x)) d x

\frac{d}{d x} (- 4.9 x^{2} + 37)

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