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

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

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

normal of

- x^{3} + 2 x^{2} - 2, at x = 1

y = - x

Solution steps

Find the tangent point:

(1, - 1)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 1

Compute the slope of the perpendicular line:

m_{p} = - 1

Find the line with slope m=

- 1

and passing through

(1, - 1) : y = - x

y = - x

\int - x^{5} e^{x} d x

\frac{\partial}{\partial x} (4 y x^{3} - 6 y^{2} x^{2})

f (x) = \frac{1}{1 + x ^{2}}, (1, \frac{1}{2})

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

\int 15 z^{3} e^{z} d z

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