What is the normal of y=2x-1+4x^2,(1,5) ?

The normal of y=2x-1+4x^2,(1,5) is y=-1/10 x+51/10

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normal of y=2x-1+4x^2,(1,5)

normal of

y = 2 x - 1 + 4 x^{2}, (1, 5)

y = - \frac{1}{10} x + \frac{51}{10}

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 5) : m = 10

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{10}

Find the line with slope m=

- \frac{1}{10}

and passing through

(1, 5) : y = - \frac{1}{10} x + \frac{51}{10}

y = - \frac{1}{10} x + \frac{51}{10}

\int k e^{- x} d x

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

g (x) = \frac{1}{4 x ^{2} + 2 x}

2 t + 3 t \cdot cos (2 t)

\frac{d}{d x} (c x e^{- x})

What is the normal of y=2x-1+4x^2,(1,5) ?
The normal of y=2x-1+4x^2,(1,5) is y=-1/10 x+51/10