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

The normal of y=x^2-x^3+x,(-2,10) is y= 1/15 x+152/15

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normal of y=x^2-x^3+x,(-2,10)

normal of

y = x^{2} - x^{3} + x, (- 2, 10)

y = \frac{1}{15} x + \frac{152}{15}

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} - x^{3} + x

(- 2, 10) : m = - 15

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{15}

Find the line with slope m=

\frac{1}{15}

and passing through

(- 2, 10) : y = \frac{1}{15} x + \frac{152}{15}

y = \frac{1}{15} x + \frac{152}{15}

(- 2.4) (7.3)

e^{2 x}

f (x) = - \frac{7}{x}, at x = - 3

f (x) = (2 sin^{3} (x) - 5 x)^{\frac{5}{3}}

(3, - 5), (7, 9)

What is the normal of y=x^2-x^3+x,(-2,10) ?
The normal of y=x^2-x^3+x,(-2,10) is y= 1/15 x+152/15