What is the normal of y=x^3-3x+5,(3,23) ?

The normal of y=x^3-3x+5,(3,23) is y=-1/24 x+185/8

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normal of y=x^3-3x+5,(3,23)

normal of

y = x^{3} - 3 x + 5, (3, 23)

y = - \frac{1}{24} x + \frac{185}{8}

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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, 23) : y = - \frac{1}{24} x + \frac{185}{8}

y = - \frac{1}{24} x + \frac{185}{8}

\frac{d}{d t} (5 sin (t) cos (t))

\int 8 ln (3 x) d x

\int_{1}^{2} ln (3 x) d x

\int \frac{sin ( 2 x )}{33 + cos ^{2} ( x )} d x

\frac{d}{d x} ((x^{2} + 3) \cdot x)

What is the normal of y=x^3-3x+5,(3,23) ?
The normal of y=x^3-3x+5,(3,23) is y=-1/24 x+185/8