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

The tangent of f(x)=-x^2-4x+2,\at x=-1 is y=-2x+3

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

tangent of

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

y = - 2 x + 3

Solution steps

Find the tangent point:

(- 1, 5)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 2

Find the line with slope m=

- 2

and passing through

(- 1, 5) : y = - 2 x + 3

y = - 2 x + 3

\frac{d}{d x} (\frac{x ^{2} - 36}{x + 6})

\frac{d}{d x} (3^{x + 10})

f (x) = 2 sin (2 x)

\int \frac{1}{x ( x + 2 )} d x

(2 y + 2 t - 7) d t + (4 y + 2 t - 1) d y = 0

What is the tangent of f(x)=-x^2-4x+2,\at x=-1 ?
The tangent of f(x)=-x^2-4x+2,\at x=-1 is y=-2x+3