What is the tangent of 4x^2+2x-1,\at x=0 ?

The tangent of 4x^2+2x-1,\at x=0 is y=2x-1

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tangent of 4x^2+2x-1,\at x=0

tangent of

4 x^{2} + 2 x - 1, at x = 0

y = 2 x - 1

Solution steps

Find the tangent point:

(0, - 1)

Find the slope of

y = 4 x^{2} + 2 x - 1 : \frac{d y}{d x} = 8 x + 2

Compute the slope of

y = 4 x^{2} + 2 x - 1

x = 0 : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 1

x y^{2} y^{^{'}} = x^{3} + 2 y^{3}

\frac{30}{s ( s + 5 ) ( s + 4 )}

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

\frac{d y}{d x} = (y e^{x})

\frac{\partial}{\partial y} (sin (2 x + y))

What is the tangent of 4x^2+2x-1,\at x=0 ?
The tangent of 4x^2+2x-1,\at x=0 is y=2x-1