What is the normal of y=x^2-9x,(3,-18) ?

The normal of y=x^2-9x,(3,-18) is y= 1/3 x-19

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normal of y=x^2-9x,(3,-18)

normal of

y = x^{2} - 9 x, (3, - 18)

y = \frac{1}{3} x - 19

Solution steps

Find the slope of

y = x^{2} - 9 x : \frac{d y}{d x} = 2 x - 9

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 3

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(3, - 18) : y = \frac{1}{3} x - 19

y = \frac{1}{3} x - 19

What is the normal of y=x^2-9x,(3,-18) ?
The normal of y=x^2-9x,(3,-18) is y= 1/3 x-19