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

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

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

tangent of

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

y = 4 x - 2

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 2) : m = 4

Find the line with slope m=

4

and passing through

(1, 2) : y = 4 x - 2

y = 4 x - 2

\int 10 (x + 1) (x^{2} + 2 x + 3)^{4} d x

\frac{d}{d x} (a r (x) cos (\frac{x + 1}{3}))

\frac{d}{d x} (18 x - \frac{2}{3} x^{3})

x \to 1 lim (e^{x} - e)

\int_{3}^{\infty} x e^{- x} d x

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