Answer to Question #93730 in Calculus for Manil

Question

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

Answer 1

$$f(x)=(x+1)(x-1)^2$$ $$f(x) \geqslant 0, x \geqslant -1; \,\, f(x) \geqslant 0, x \geqslant -1$$ $$f(x)=0 \Rightarrow x=-1, \,\, x=1$$ $$f'(x)=3x^2-2x-1 \Rightarrow x=-1/3, \,\, x=1$$ $$f \uparrow, x<-1/3, x>1; \,\, f \downarrow -1/3<x<1$$ $$y^2=f(x) \Rightarrow y=\pm \sqrt{f(x)}$$

$\sqrt{f(x)}$ have the same zeros as $f(x)$ and it determining for $x \geqslant -1$. Moreover, $y'=-\frac{f'(x)}{2\sqrt{f(x)}}$ so we have $y \uparrow, -1<x<-1/3, x>1; \,\, y \downarrow -1/3<x<1$. And we see that at $x=-1: \,\, y' \to \infty$, so we have vertical tangent. And at $x \to 1 \pm: \,\,y'= \pm \sqrt{2}$, i.e. parts of function have $90^{\circ}$ angle between each other and $45^{\circ}$ and $135^{\circ}$ between $x$-axis (from right to left).

$y=-\sqrt{f(x)}$ is given by symmetric mapping of $y=\sqrt{f(x)}$ with respect to $x$-axis.