What is the tangent of f(x)=x^2+1,(2,5) ?

The tangent of f(x)=x^2+1,(2,5) is y=4x-3

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tangent of f(x)=x^2+1,(2,5)

tangent of

f (x) = x^{2} + 1, (2, 5)

y = 4 x - 3

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

(2, 5) : m = 4

Find the line with slope m=

4

and passing through

(2, 5) : y = 4 x - 3

y = 4 x - 3

\int \frac{2}{1 + 9 x ^{2}} d x

(x^{2} - 1)^{2}

\int x ln (5 + x) d x

\frac{\partial}{\partial y} (3 x^{3} y)

\int sin (x) \cdot 2 x d x

What is the tangent of f(x)=x^2+1,(2,5) ?
The tangent of f(x)=x^2+1,(2,5) is y=4x-3