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

The normal of y=x^3-3x+3,(2,5) is y=-1/9 x+47/9

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

normal of

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

y = - \frac{1}{9} x + \frac{47}{9}

Solution steps

Find the slope of

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

Compute the slope of

y = x^{3} - 3 x + 3

(2, 5) : m = 9

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{9}

Find the line with slope m=

- \frac{1}{9}

and passing through

(2, 5) : y = - \frac{1}{9} x + \frac{47}{9}

y = - \frac{1}{9} x + \frac{47}{9}

\frac{d}{d x} (sin (0))

\frac{d}{d x} (\frac{1}{( x - 1 ) ^{2}})

\frac{4 x ^{2} + 2 x + 4}{x}

\frac{d y}{d x} = e^{4 x} + 5 y

What is the normal of y=x^3-3x+3,(2,5) ?
The normal of y=x^3-3x+3,(2,5) is y=-1/9 x+47/9