What is the tangent of f(x)=x^2+1,(3,10) ?

The tangent of f(x)=x^2+1,(3,10) is y=6x-8

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tangent of f(x)=x^2+1,(3,10)

tangent of

f (x) = x^{2} + 1, (3, 10)

y = 6 x - 8

Solution steps

Find the slope of

f (x) = x^{2} + 1 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 1

(3, 10) : m = 6

Find the line with slope m=

6

and passing through

(3, 10) : y = 6 x - 8

y = 6 x - 8

f (x) = 30 x

f (x) = ln (\frac{x ^{2}}{x ^{2} + 4})

\int \frac{1}{( 15 - x ) ( 60 - x )} d x

x \to - 1 lim (\frac{k x + 3}{4})

y = x^{3} - 3 x^{2} + 4

What is the tangent of f(x)=x^2+1,(3,10) ?
The tangent of f(x)=x^2+1,(3,10) is y=6x-8