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

The normal of y= 1/3 x^3-2x+2,(-3,-1) is y=-1/7 x-10/7

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

normal of

y = \frac{1}{3} x^{3} - 2 x + 2, (- 3, - 1)

y = - \frac{1}{7} x - \frac{10}{7}

Solution steps

Find the slope of

y = \frac{1}{3} x^{3} - 2 x + 2 : \frac{d y}{d x} = x^{2} - 2

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 = 7

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{7}

Find the line with slope m=

- \frac{1}{7}

and passing through

(- 3, - 1) : y = - \frac{1}{7} x - \frac{10}{7}

y = - \frac{1}{7} x - \frac{10}{7}

\int \frac{x - 3}{x ^{2} - 6 x + 10} d x

\frac{d}{d x} (\frac{x y}{x - y})

\frac{d}{d x} (3 x - \frac{5}{x})

y = \frac{1}{2} x^{2}, 2 x

\frac{d}{d x} (5 cos (2 x))

What is the normal of y= 1/3 x^3-2x+2,(-3,-1) ?
The normal of y= 1/3 x^3-2x+2,(-3,-1) is y=-1/7 x-10/7