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

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

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normal of y=(x^2)/(x-1),(0,0)

normal of

y = \frac{x ^{2}}{x - 1}, (0, 0)

x = 0

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{x ^{2}}{x - 1}

(0, 0) : m = 0

Compute the slope of the perpendicular line:

m_{p} = \infty

Find the line with slope m=

\infty

and passing through

(0, 0) : x = 0

x = 0

f (x) = (x^{2} - 3 x) ln (x^{2} - 3 x)

z = - 3 - 3 i

y = x^{2} + 1

f (x) = - 2 x + 2

ln (3) x^{2}

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