What is the normal of y=x^2-5x,(5,0) ?

The normal of y=x^2-5x,(5,0) is y=-1/5 x+1

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normal of y=x^2-5x,(5,0)

normal of

y = x^{2} - 5 x, (5, 0)

y = - \frac{1}{5} x + 1

Solution steps

Find the slope of

y = x^{2} - 5 x : \frac{d y}{d x} = 2 x - 5

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 = 5

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{5}

Find the line with slope m=

- \frac{1}{5}

and passing through

(5, 0) : y = - \frac{1}{5} x + 1

y = - \frac{1}{5} x + 1

What is the normal of y=x^2-5x,(5,0) ?
The normal of y=x^2-5x,(5,0) is y=-1/5 x+1