What is the normal of y=5x^3-3x,(1,2) ?

The normal of y=5x^3-3x,(1,2) is y=-1/12 x+25/12

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normal of y=5x^3-3x,(1,2)

normal of

y = 5 x^{3} - 3 x, (1, 2)

y = - \frac{1}{12} x + \frac{25}{12}

Solution steps

Find the slope of

y = 5 x^{3} - 3 x : \frac{d y}{d x} = 15 x^{2} - 3

Compute the slope of

y = 5 x^{3} - 3 x

(1, 2) : m = 12

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{12}

Find the line with slope m=

- \frac{1}{12}

and passing through

(1, 2) : y = - \frac{1}{12} x + \frac{25}{12}

y = - \frac{1}{12} x + \frac{25}{12}

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

x \to \infty lim (24 x^{7} e^{- x^{8}}, [0, x])

\frac{\partial}{\partial y} (x e^{3 x y})

\frac{s}{s ^{2} + 4 s - 5}

x^{2} (3 x^{2} - x - 14)

What is the normal of y=5x^3-3x,(1,2) ?
The normal of y=5x^3-3x,(1,2) is y=-1/12 x+25/12