What is the normal of y=x^3-3x+3,(3,21) ?

The normal of y=x^3-3x+3,(3,21) is y=-1/24 x+169/8

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normal of y=x^3-3x+3,(3,21)

normal of

y = x^{3} - 3 x + 3, (3, 21)

y = - \frac{1}{24} x + \frac{169}{8}

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

(3, 21) : m = 24

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{24}

Find the line with slope m=

- \frac{1}{24}

and passing through

(3, 21) : y = - \frac{1}{24} x + \frac{169}{8}

y = - \frac{1}{24} x + \frac{169}{8}

cos (4 x)

csc (x) (x + cot (x))

e^{- x^{2}}

\int_{0}^{s i n (x)} 6 t d t

y^{^{''}} + 2 y^{^{'}} - 48 y = 0

What is the normal of y=x^3-3x+3,(3,21) ?
The normal of y=x^3-3x+3,(3,21) is y=-1/24 x+169/8