What is the normal of y=x^2-2x,(5,15) ?

The normal of y=x^2-2x,(5,15) is y=-1/8 x+125/8

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normal of y=x^2-2x,(5,15)

normal of

y = x^{2} - 2 x, (5, 15)

y = - \frac{1}{8} x + \frac{125}{8}

Solution steps

Find the slope of

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

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 = 8

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{8}

Find the line with slope m=

- \frac{1}{8}

and passing through

(5, 15) : y = - \frac{1}{8} x + \frac{125}{8}

y = - \frac{1}{8} x + \frac{125}{8}

\frac{1}{( s + 2 ) ^{2}}

\int \frac{sin ( x )}{cos ( x ) + 1} d x

(y - y x^{2}) \frac{d y}{d x} = (y + 1)^{2}

f (x) = 24

\frac{\partial}{\partial y} (6 x + 5 y)

What is the normal of y=x^2-2x,(5,15) ?
The normal of y=x^2-2x,(5,15) is y=-1/8 x+125/8