What is the normal of y=-2x^3+6/x ,\at x=-1 ?

The normal of y=-2x^3+6/x ,\at x=-1 is y= 1/12 x-47/12

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normal of y=-2x^3+6/x ,\at x=-1

normal of

y = - 2 x^{3} + \frac{6}{x}, at x = - 1

y = \frac{1}{12} x - \frac{47}{12}

Solution steps

Find the tangent point:

(- 1, - 4)

Find the slope of

y = - 2 x^{3} + \frac{6}{x} : \frac{d y}{d x} = - 6 x^{2} - \frac{6}{x ^{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 = - 12

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{12}

Find the line with slope m=

\frac{1}{12}

and passing through

(- 1, - 4) : y = \frac{1}{12} x - \frac{47}{12}

y = \frac{1}{12} x - \frac{47}{12}

\int \frac{x ^{3}}{2 x ^{4} + x ^{2} - 1} d x

\frac{d}{d x} (\frac{x ^{4} + 3}{3 x})

\frac{d}{d x} (\frac{x ^{2}}{8 - 8 x})

\int x^{2} (x^{3} + 1) d x

x \to 0 lim (sin (x) + cos (x))

What is the normal of y=-2x^3+6/x ,\at x=-1 ?
The normal of y=-2x^3+6/x ,\at x=-1 is y= 1/12 x-47/12