What is the tangent of f(x)=x^2+4,(3,13) ?

The tangent of f(x)=x^2+4,(3,13) is y=6x-5

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tangent of f(x)=x^2+4,(3,13)

tangent of

f (x) = x^{2} + 4, (3, 13)

y = 6 x - 5

Solution steps

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

(3, 13) : m = 6

Find the line with slope m=

6

and passing through

(3, 13) : y = 6 x - 5

y = 6 x - 5

\int e^{- x} d x

\frac{\partial}{\partial x} (ln (2))

32 - 2 x^{2}, [0, 7]

x \to 0 lim (\frac{( x ^{3} )}{tan ^{3} ( 2 x )})

\int_{1}^{5} \frac{1}{x ^{2}} d x

What is the tangent of f(x)=x^2+4,(3,13) ?
The tangent of f(x)=x^2+4,(3,13) is y=6x-5