What is the normal of y=x^2-5x,(7,14) ?

The normal of y=x^2-5x,(7,14) is y=-1/9 x+133/9

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normal of y=x^2-5x,(7,14)

normal of

y = x^{2} - 5 x, (7, 14)

y = - \frac{1}{9} x + \frac{133}{9}

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 = 9

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{9}

Find the line with slope m=

- \frac{1}{9}

and passing through

(7, 14) : y = - \frac{1}{9} x + \frac{133}{9}

y = - \frac{1}{9} x + \frac{133}{9}

x \to - \infty lim (x^{3} + 2)

2 x + y

x \to 6 - lim (\frac{7}{x - 6})

(4 x^{2} + 3 y^{2}) d x - (2 x y) d y = 0

y = x^{2} - 4

What is the normal of y=x^2-5x,(7,14) ?
The normal of y=x^2-5x,(7,14) is y=-1/9 x+133/9