What is the normal of y=4x^3-9x^2-12x+12,(1,-5) ?

The normal of y=4x^3-9x^2-12x+12,(1,-5) is y= 1/18 x-91/18

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normal of y=4x^3-9x^2-12x+12,(1,-5)

normal of

y = 4 x^{3} - 9 x^{2} - 12 x + 12, (1, - 5)

y = \frac{1}{18} x - \frac{91}{18}

Solution steps

Find the slope of

y = 4 x^{3} - 9 x^{2} - 12 x + 12 : \frac{d y}{d x} = 12 x^{2} - 18 x - 12

Compute the slope of

y = 4 x^{3} - 9 x^{2} - 12 x + 12

(1, - 5) : m = - 18

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{18}

Find the line with slope m=

\frac{1}{18}

and passing through

(1, - 5) : y = \frac{1}{18} x - \frac{91}{18}

y = \frac{1}{18} x - \frac{91}{18}

\frac{d}{d x} (\frac{tan ( x ) - 1}{sec ( x )})

\frac{d}{d x} (2 arccos (\frac{x}{4}))

f (x) = 2 sin (x), at x = \frac{π}{3}

\int x^{2} 6 x + 5 d x

f (x) = 3 a^{3 x}

What is the normal of y=4x^3-9x^2-12x+12,(1,-5) ?
The normal of y=4x^3-9x^2-12x+12,(1,-5) is y= 1/18 x-91/18