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

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

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

tangent of

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

y = 4 x + 4

Solution steps

Find the tangent point:

(2, 12)

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

x = 2 : m = 4

Find the line with slope m=

4

and passing through

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

y = 4 x + 4

\int e^{x} + 2 d x

x \to - \infty lim (x^{2} + 6)

\frac{d}{d x} ((x^{4} + 3 x^{2} - 2)^{5})

\int \frac{( 3 y - 8 )}{y ^{2} - 4 y - 5} d y

\int \frac{1}{( 3 x + 7 ) ^{2}} d x

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