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

The normal of y=-x^2-2x+2,(-3,-1) is y=-1/4 x-7/4

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

normal of

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

y = - \frac{1}{4} x - \frac{7}{4}

Solution steps

Find the slope of

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

Compute the slope of

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

(- 3, - 1) : m = 4

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{4}

Find the line with slope m=

- \frac{1}{4}

and passing through

(- 3, - 1) : y = - \frac{1}{4} x - \frac{7}{4}

y = - \frac{1}{4} x - \frac{7}{4}

sin (x), 0

y^{^{'}} - \frac{9}{x} y = \frac{y ^{5}}{x ^{6}}

n = 1 \sum \infty e^{- 2 n}

x y^{^{'}} = y^{2} + y

\frac{d}{d x} ((4 x^{2} - 5 x) e^{x})

What is the normal of y=-x^2-2x+2,(-3,-1) ?
The normal of y=-x^2-2x+2,(-3,-1) is y=-1/4 x-7/4