What is the normal of y=x^2-8x,(5,-15) ?

The normal of y=x^2-8x,(5,-15) is y=-1/2 x-25/2

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normal of y=x^2-8x,(5,-15)

normal of

y = x^{2} - 8 x, (5, - 15)

y = - \frac{1}{2} x - \frac{25}{2}

Solution steps

Find the slope of

y = x^{2} - 8 x : \frac{d y}{d x} = 2 x - 8

Compute the slope of

y = x^{2} - 8 x

(5, - 15) : m = 2

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(5, - 15) : y = - \frac{1}{2} x - \frac{25}{2}

y = - \frac{1}{2} x - \frac{25}{2}

\int cos (r) d r

x \cdot y^{^{'}} = 2 y

t \to 2 lim (\frac{t ^{3} - 8}{t - 2})

\frac{d}{d x} (3 x^{8})

x \to - 1 lim (4 x^{2} + 6)

What is the normal of y=x^2-8x,(5,-15) ?
The normal of y=x^2-8x,(5,-15) is y=-1/2 x-25/2