Draw the graph of the relation `y^(2)=x^(5)(2-x)`

Explanation

We have `y^(2)=x^(5)(2-x)`
`therefore y =+-sqrt(2x-x^(2))`
Let us first draw the graph of `y=f(x) =x ^(2)sqrt(2x-x^(2))`.
Clearly, the domain of function is `[0,2]`.
So, the graph meets the x-axis at (0,0) and (2,0).
Also `f(x) ge0`
`f^(')(x) = 2xsqrt(2x-x^(2))+ x^(2)(1-x)/sqrt(2x-x^(2)) = (x^(2)(5-3x))/sqrt(2x-x^(2))`
`f^(')(x)=0` `therefore x=5//3`, which is the point of maxima.
`f^('')(x) = ((10x-9x^(2))sqrt(2x-x^(2))-(1-x)/sqrt(2x-x^(2))(5x^(2)-3x^(2)))/(2x-x^(2))`
`x^(2)(6x^(2)-20x+15)/((2x-x^(2)sqrt(2x-x^(2))))`
`f^('')(x)=0 therefore x=(10-sqrt(10))/6`, which is the point of inflection.
Hence, the graph of `y=f(x)` is as shown in the following figure.

To draw the graph of `y=-x^(2)sqrt(2x-x^(2))`, reflect the above graph in the x-axis.
Hence, the graph of the given relation `y^(2)=x^(5)(2-x)` is as shown in the following figure.