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

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

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

tangent of

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

y = - 8 x - 4

Solution steps

Find the tangent point:

(- 2, 12)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = - 8

Find the line with slope m=

- 8

and passing through

(- 2, 12) : y = - 8 x - 4

y = - 8 x - 4

\frac{d}{d x} ((3 x^{2} + 2 x) (x^{4} - 3 x + 1))

y^{^{'''}} + y^{^{'}} = sec (x)

\frac{\partial}{\partial x} (x^{2 y})

sec (x) \frac{d y}{d x} = e^{y}

\int cos^{3} (x) (1 + sin (x)) d x

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