What is the tangent of f(x)=x^2+8,(2,12) ?

The tangent of f(x)=x^2+8,(2,12) is y=4x+4

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tangent of f(x)=x^2+8,(2,12)

tangent of

f (x) = x^{2} + 8, (2, 12)

y = 4 x + 4

Solution steps

Find the slope of

f (x) = x^{2} + 8 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 8

(2, 12) : m = 4

Find the line with slope m=

4

and passing through

(2, 12) : y = 4 x + 4

y = 4 x + 4

\int 3 x^{\frac{2}{3}} d x

- \frac{3}{x ^{3}}

\int (\frac{d y}{1})^{2} d x

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

\frac{\partial}{\partial y} (x^{2} y z^{2} + sin (yz))

What is the tangent of f(x)=x^2+8,(2,12) ?
The tangent of f(x)=x^2+8,(2,12) is y=4x+4