Answer to Question #113935 in Calculus for Bibhuti bhusan dehury

Question

Tracing of curve y^2=x(x-2)^2

Answer 1

Given curve $y^2 = x(x-2)^2$.

At $x=0 \hspace{0.05 in} and \hspace{0.05 in} x= 2, y = 0.$

Also $y^2 \geq 0$ for all value of x.

Given curve is symmetric about x-axis as by putting y = -y curve remain unchanged.

So, let discuss only positive side of y-axis. Then Negative side of y-axis plot symmetrically around x-axis.

So $y = \sqrt{x} (x-2)$.

Now, Differentiate given curve with respect to x, we get

$y' = \sqrt{x} + \frac{x-2}{2\sqrt{x}}$

So, $y' > 0$ for $x > 2$. Hence, y is an increasing function after x=2.

Now $y' = 0 \implies 2x+x-2 = 0 \implies x = 2/3 = 0.66.$

And $y'>0$ for $0 \leq x < 0.66$ , $y' <0$ for $0.66<x<2$.