Answer to Question #92776 in Calculus for MGM

Question

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

Answer 1

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

Solution:

Step 1: Domain

$D(y): (x+1)(x-1)^2\geqslant0$

$(x+1)\geqslant0$

$x\geqslant-1$

$D(y)=[-1;+\infin)$

Step 2: Range

$E(y)=\R$

Step 3: Symmetry

$(-y)^{2}=(x+1)(x-1)^{2}=y^{2}$

The curve is symmetric about x-axis

Step 4: Asymptotes

It has no asymptotes

Step 5: Intersection points with axes:

x-axis: $y=0, x=\pm1$

y-axis: $x=0, y=\pm1$

Step 6: Monotonicity

$y=\pm(x-1)\sqrt{x+1}$

$y'=\pm(\sqrt{x+1}+\frac{x-1}{2\sqrt{x+1}})$

$y'=\pm\frac{2x+2+x-1}{2\sqrt{x+1}}$

$y'=\pm\frac{3x+1}{2\sqrt{x+1}}$

Let $y'=0$

$\pm\frac{3x+1}{2\sqrt{x+1}}=0$

If $x=-1$ then $y'=\infin$ or $y'$ does not exists

$\pm(3x+1)=0$

$x=-\frac{1}{3}$

$y^{2}(-\frac{1}{3})=\frac{2}{3}\cdot\frac{16}{9}=\frac{32}{27}$

$y=\pm\frac{\sqrt{32}}{\sqrt{27}}\approx\pm1.09$

Let $y<0$

So that$(-\frac{1}{3};-\frac{\sqrt{32}}{\sqrt{27}})$ is minimum

Let $y>0$

So that $(-\frac{1}{3};\frac{\sqrt{32}}{\sqrt{27}})$ is maximum

Step 7: Curve