Question

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

Answer 1

Explanation

Since the equation does not change when we replace y with –y, then the curve has reflective symmetry with x-axis.

We can find intersection points of the curve with x-axis from equation y = 0. Hence, (x+1)(x−1)² = 0, or x = -1, x = 1. From equation x = 0 we find intersection points of the curve with y-axis: y² = 1, or y = -1, y = 1.

Because x+1 = y²/(x-1)² (x≠1), then x+1≥0, x≥ -1.

Let us rewrite the equation

y²-(x+1) (x−1)² = 0, that is (y-(x-1)$\sqrt{x+1}$ )(y+(x-1)$\sqrt{x+1}$ ) = 0.

Hence the curve has two branches y=(x-1)$\sqrt{x+1}$, y=-(x-1)$\sqrt{x+1}$ .

Because y/x=(1-1/x)$\sqrt{x+1}$ →$\infin$ as x tends $\infty$, these branches have not asymptotes.

A branch y=(x-1)$\sqrt{x+1}$ is the graph of the function y=(x-1)$\sqrt{x+1}$. For analyzing this function we find its derivative. It is equal to y'=$(3x+1)\over2\sqrt{x+1}$. The derivative is negative for x<-1/3, it is positive for x>-1/3 and equals to zero at x=-1/3. Hence the function is decreasing over the interval ]-1, -1/3[ and is increasing over the interval ]-1/3,$\infin$ [. At the point x=-1/3 it has the minimal value y_min = $-4\sqrt{2}\over3\sqrt{3}$.

Since the second derivative of the function y'' = $\sqrt{x+1}(3x+5)\over(2x+2)^2$ is positive over the interval ]-1,$\infin$[, then the graph of function is concave up on the interval ]-1,$\infin$ [.

Using the mentioned properties we can trace the curve.