What is the tangent of y=-2x^2-4x+1,(0,1) ?

The tangent of y=-2x^2-4x+1,(0,1) is y=-4x+1

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tangent of y=-2x^2-4x+1,(0,1)

tangent of

y = - 2 x^{2} - 4 x + 1, (0, 1)

y = - 4 x + 1

Solution steps

Find the slope of

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

Compute the slope of

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

(0, 1) : m = - 4

Find the line with slope m=

- 4

and passing through

(0, 1) : y = - 4 x + 1

y = - 4 x + 1

\frac{x}{2 x - 1}

\frac{d y}{d x} = x y + 3 x + 4 y + 12

\int ln (3 x - 2) d x

\int x^{3} + sin (x) d x

\frac{d}{d x} (\frac{sin ( x )}{x})

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