What is the tangent of f(x)=x^2+5,\at x=5 ?

The tangent of f(x)=x^2+5,\at x=5 is y=10x-20

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tangent of f(x)=x^2+5,\at x=5

tangent of

f (x) = x^{2} + 5, at x = 5

y = 10 x - 20

Solution steps

Find the tangent point:

(5, 30)

Find the slope of

f (x) = x^{2} + 5 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 5

x = 5 : m = 10

Find the line with slope m=

10

and passing through

(5, 30) : y = 10 x - 20

y = 10 x - 20

x \to 3 lim (2)

\frac{d}{d x} (π)

n = 1 \sum \infty n^{- \frac{1}{9}}

\frac{d}{d x} (\frac{x ^{2}}{( x - y ) ^{2}})

x \to 9 lim (\frac{( x + 2 )}{2})

What is the tangent of f(x)=x^2+5,\at x=5 ?
The tangent of f(x)=x^2+5,\at x=5 is y=10x-20