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

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

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

tangent of

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

y = 2 x - 8

Solution steps

Find the tangent point:

(2, - 4)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 8

\frac{\partial}{\partial t} (sin (x - a t))

cos (2 t) (1 + e^{- 3 t})

\frac{d}{d x} ((x^{3} + 1)^{5})

\frac{\partial}{\partial x} (e^{- \frac{1}{x}})

- 3 x^{2} y + y^{^{'}} = - 9 x^{2}

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