What is the normal of y=x^2-4x,(7,21) ?

The normal of y=x^2-4x,(7,21) is y=-1/10 x+217/10

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normal of y=x^2-4x,(7,21)

normal of

y = x^{2} - 4 x, (7, 21)

y = - \frac{1}{10} x + \frac{217}{10}

Solution steps

Find the slope of

y = x^{2} - 4 x : \frac{d y}{d x} = 2 x - 4

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 = 10

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{10}

Find the line with slope m=

- \frac{1}{10}

and passing through

(7, 21) : y = - \frac{1}{10} x + \frac{217}{10}

y = - \frac{1}{10} x + \frac{217}{10}

What is the normal of y=x^2-4x,(7,21) ?
The normal of y=x^2-4x,(7,21) is y=-1/10 x+217/10