Answer to Question #95459 in Calculus for Amaljith

Question

Trace the curve y²=(x+1)(x-1)² by showing all the properties you use to trace it.

Answer 1

$y_1=(x-1)\sqrt{(x+1)}\\ y_2=-(x-1)\sqrt{x+1}$

$y_1'=\sqrt{(x+1)}+\frac{(x-1)}{2\sqrt{(x+1)}}=\frac{3x+1}{2\sqrt{(x+1)}}\\ y_1''=\frac{3}{2\sqrt{(x+1)}}-\frac{3x+1}{4(x+1)\sqrt{(x+1)}}=\\ \frac{3x+5}{4(x+1)\sqrt{(x+1)}}\\ y_2'=-\frac{3x+1}{2\sqrt{(x+1)}}\\ y_2''=-\frac{3x+5}{4(x+1)\sqrt{(x+1)}}$

zero of $y_1'=0\ and \ y_2'=0\,is \, x=-1/3$

Since $y_1''(-1/3)>0,\ y_1$ has minimum at this point.

Since $y_2''(-1/3)<0,\ y_2\$ has maximum at this point.

Since $0\le y^2 \ and \ 0\le(x-1)^2$ ,it follows that $0\le(x+1)\ and\ -1\le x$

Intersections with axis:

$y=0, \ (x+1)(x-1)²=0,\ x=\pm1$

$x=0,\ y=\pm 1$