What is the normal of y=x^2-8x,(3,-15) ?

The normal of y=x^2-8x,(3,-15) is y= 1/2 x-33/2

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normal of y=x^2-8x,(3,-15)

normal of

y = x^{2} - 8 x, (3, - 15)

y = \frac{1}{2} x - \frac{33}{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

(3, - 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

(3, - 15) : y = \frac{1}{2} x - \frac{33}{2}

y = \frac{1}{2} x - \frac{33}{2}

\int \frac{1}{x ^{3} + 8 x} d x

f (x) = 4 cos (2 x)

n = 0 \sum \infty \frac{4}{n}

y = tan^{2} (x), (\frac{π}{4}, 1)

\frac{d}{d x} (e^{\frac{1}{2} x^{2}})

What is the normal of y=x^2-8x,(3,-15) ?
The normal of y=x^2-8x,(3,-15) is y= 1/2 x-33/2