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

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

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

tangent of

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

y = 4 x - 8

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 4

Find the line with slope m=

4

and passing through

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

y = 4 x - 8

f (x) = \frac{6 x}{x ^{2} + 1}, at x = 0

\frac{d}{d x} (\frac{π}{x} + ln (2))

\frac{\partial}{\partial x} ((x)^{2} (y)^{4})

\frac{x}{x + 5}

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

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