What is the tangent of f(x)=x^2+3x-1,\at x=0 ?

The tangent of f(x)=x^2+3x-1,\at x=0 is y=3x-1

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tangent of f(x)=x^2+3x-1,\at x=0

tangent of

f (x) = x^{2} + 3 x - 1, at x = 0

y = 3 x - 1

Solution steps

Find the tangent point:

(0, - 1)

Find the slope of

f (x) = x^{2} + 3 x - 1 : \frac{df ( x )}{d x} = 2 x + 3

Compute the slope of

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

x = 0 : m = 3

Find the line with slope m=

3

and passing through

(0, - 1) : y = 3 x - 1

y = 3 x - 1

x \to 0 - lim (x \cdot ln (x))

\frac{d}{d x} (\frac{6 x + y}{x ^{2} + y ^{2}})

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

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

f (x) = e^{x} - e^{- x}

What is the tangent of f(x)=x^2+3x-1,\at x=0 ?
The tangent of f(x)=x^2+3x-1,\at x=0 is y=3x-1