What is the normal of y=5x^3+4x^2-6,(1,3) ?

The normal of y=5x^3+4x^2-6,(1,3) is y=-1/23 x+70/23

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normal of y=5x^3+4x^2-6,(1,3)

normal of

y = 5 x^{3} + 4 x^{2} - 6, (1, 3)

y = - \frac{1}{23} x + \frac{70}{23}

Solution steps

Find the slope of

y = 5 x^{3} + 4 x^{2} - 6 : \frac{d y}{d x} = 15 x^{2} + 8 x

Compute the slope of

y = 5 x^{3} + 4 x^{2} - 6

(1, 3) : m = 23

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{23}

Find the line with slope m=

- \frac{1}{23}

and passing through

(1, 3) : y = - \frac{1}{23} x + \frac{70}{23}

y = - \frac{1}{23} x + \frac{70}{23}

\int e^{- 2 x} \cdot sin (x) d x

x \to 2 lim ((- 2)^{3})

\frac{\partial}{\partial x} (\frac{x y}{( x + y )})

\frac{d}{d x} (- \frac{1}{1 + x ^{2}})

\frac{d}{d y} (4 y^{2})

What is the normal of y=5x^3+4x^2-6,(1,3) ?
The normal of y=5x^3+4x^2-6,(1,3) is y=-1/23 x+70/23