Question

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

Answer 1

1. Symmetry 

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

All powers of $y$ in the equation are even. 

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

Therefore, curve is symmetrical about $x-$ axis.

Curve is not symmetrical about $y-$ axis.

Curve is not symmetrical about the line $y=x.$

Curve is not symmetrical about the line $y=-x$

Curve is not symmetrical in opposite quadrants.

2. Regions where no Part of the curve lies 

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

Since $y^2\geq0,$ we have $(x+1)(x-1)^2\geq0 =>x\geq-1$

Therefore, $x\geq-1, y\isin\Reals$

3. Origin

$$0^2=(0+1)(0-1)^2$$ $$0=1, False$$

A curve does not pass through the origin.

4. Intercepts

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

Intersection with $x-$ axis: Put $y=0$ in the equation of the curve and solve the resulting equation

$$0^2=(x+1)(x-1)^2=>x_1=-1, x_2=1$$

Intersection with $x-$ axis:$Point(-1, 0), Point(1,0)$

Intersection with $y-$axis: Put $x=0$ in the equation of the curve and solve the resulting equation

$$y^2=(0+1)(0-1)^2=>y_1=-1, y_2=1$$

Intersection with $y-$axis: $Point(0, -1), Point(0,1)$

5. Tangent to the curve

In order to find the tangent(s) at the point (1, 0), we should shift the origin to (1, 0) and then the tangent(s) at this new origin will be obtained by equating to zero the lowest degree term.

$$\widetilde{y}^2=\widetilde{x}^2(\widetilde{x}+2)$$ $$\widetilde{y}=\pm\sqrt{2}$$

The point (1,0) is an ordinary double point; around it the curve has two branches. The intersection point (1,0) where the graph crosses itself is called a node since two branches of the curve have distinct tangents.  

6. Asymptotes

The equations of a vertical asymptotes are obtained by equating the coefficient of highest degree term in y to zero if it is not a constant. 

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

The coefficient of highest degree term in y, the coefficient of $y^2$ is 1 , which is a constant .

There is no vertical asymptote.

The coefficient of highest degree term in x, the coefficient of $x^3$ is 1 , which is a constant . There is no horizontal asymptote.

To obtain the equations of oblique asymptotes, substitutes y = mx + c in the given equation. Then equate the coefficients of the highest degree term in x and next highest degree term in x to zero, if it is not a constant, to determine m and c. If the values of m and c exists, then y = mx + c is the equation of the oblique asymptote. 

$$(mx+c)^2=(x+1)(x-1)^2$$ $$m^2x^2+2mcx+c^2=x^3-x^2-x+1$$

The coefficient of highest degree term in x, the coefficient of $x^3$ is 1 , which is a constant .

There is no oblique asymptote.

7. First derivative

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

$$2y{dy \over dx}=(x-1)^2+2(x+1)(x-1)$$ $${dy \over dx}={(x-1)(3x+1) \over y}$$

If y>0:

$${dy \over dx}>0, \ when\ -1<x<-1/3 \ or \ x>1, y \ \ increases$$ $${dy \over dx}<0, \ when\ -1/2<x<1, y \ \ decreases$$

If y<0:

$${dy \over dx}>0, \ when\ -1/2<x<1, y \ \ increases$$ $${dy \over dx}<0, \ when\ -1<x<-1/3 \ or \ x>1, y \ \ decreases$$

8. Graph