What is the tangent of f(x)=x^2+2,(2,6) ?

The tangent of f(x)=x^2+2,(2,6) is y=4x-2

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tangent of f(x)=x^2+2,(2,6)

tangent of

f (x) = x^{2} + 2, (2, 6)

y = 4 x - 2

Solution steps

Find the slope of

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

Compute the slope of

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

(2, 6) : m = 4

Find the line with slope m=

4

and passing through

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

y = 4 x - 2

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

y^{^{'}} = y ln (\frac{450}{y})

f (x) = 4 cos (4 x)

\int \frac{1}{e ^{- y} + 1} d y

\frac{d}{d x} (\frac{x}{1 - ln ( x - 6 )})

What is the tangent of f(x)=x^2+2,(2,6) ?
The tangent of f(x)=x^2+2,(2,6) is y=4x-2