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

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

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

normal of

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

y = - x - 2

Solution steps

Find the slope of

y = 2 x^{3} - 5 x : \frac{d y}{d x} = 6 x^{2} - 5

Compute the slope of

y = 2 x^{3} - 5 x

(1, - 3) : m = 1

Compute the slope of the perpendicular line:

m_{p} = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x - 2

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

\frac{\partial}{\partial x} (\frac{1}{2} sin (2 x - y))

\frac{\partial}{\partial x} (\frac{x ^{3} b ^{4}}{c ^{2} d ^{4}})

y = 4 x, x = 0, y = 4

\frac{d y}{d x} = \frac{( y - x y )}{x ^{2}}

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