Question

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

Answer 1

The curve $y^2 = (x +1) (x −1)^2$ is well defined for

$(x +1) (x −1)^2\geq 0 \Rightarrow (x +1)\geq 0 \Rightarrow x \geq -1 \\$  

All powers of 𝑦 in the equation are even

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

Therefore, curve is symmetrical about 𝑥- axis.

A curve does not pass through the origin:

A curve does not pass through the origin:

$0^2 = (0 +1) (0 −1)^2 \\ 0=1- False$

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

$0^2 = (x +1) (x −1)^2$

$(x +1)=0 \Rightarrow x_1=-1\\ (x −1)^2=0 \Rightarrow x_2=1$

Intersection with 𝑥- axis: 𝑃𝑜𝑖𝑛𝑡s (-1, 0) and (-1, 0).

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

$y^2 = (0 +1) (0 −1)^2\\ y^2 =1*1 \Rightarrow y=\pm 1.$

Intersection with y- axis: 𝑃𝑜𝑖𝑛𝑡s (0,-1) and (0,1).

$y^2 = (x +1) (x −1)^2$ , $x \geq -1$.

So,

$y = \pm\sqrt{(x +1) (x −1)^2}$ .

Consider the case

$y = \sqrt{(x +1) (x −1)^2}$ (1)

$y = \sqrt{(x +1)} |x −1|$

$y= \begin{cases} (x −1)\sqrt{(x +1)} , &\text{if } x\geq 1\\ - (x −1)\sqrt{(x +1)}, &\text{if } -1\leq x< 1 \end{cases}$

First derivative:

$\frac {dy}{ dx}= \begin{cases} \sqrt{(x +1)}+ \frac {(x −1)}{ 2\sqrt{(x +1)}} , &\text{if } x\geq 1\\ - (\sqrt{(x +1)}+ \frac {(x −1)}{ 2\sqrt{(x +1)}}), &\text{if } -1 < x< 1 \end{cases}$

$\frac {dy}{ dx}= \begin{cases} \frac {3x +1}{ 2\sqrt{x +1}} , &\text{if } x\geq 1\\ - \frac {3x +1}{ 2\sqrt{x +1}}, &\text{if } -1\leq x< 1 \end{cases}$

If $x =-\frac {1}{3}, \frac {𝑑𝑦}{𝑑𝑥} = 0,$

If $x> -\frac {1}{3}, \frac {𝑑𝑦}{𝑑𝑥} < 0,$  y -decreases,

If $-1\leq x< -\frac {1}{3}, \frac {𝑑𝑦}{𝑑𝑥} > 0$, y -increases,

$x \geq 1 , \frac {𝑑𝑦}{𝑑𝑥} > 0$ - y -increases,

In point x = -1/3, the derivative of the function changes the sign from (+) to (-). Therefore, the point x = -1/3 is the maximum point.

 Second derivative

$\frac {d^2y}{ dx^2}= -\frac {3x+5}{ 4(x+1)^ \frac {3}{ 4}}$

$\frac {d^2y}{ dx^2}=\begin{cases} \frac {3x+5}{ 4(x+1)^ \frac {3}{ 2}} , &\text{if } x\geq 1\\ - \frac {3x+5}{ 4(x+1)^ \frac {3}{2}}, &\text{if } -1< x< 1 \end{cases}$

$\frac {d^2y}{ dx^2} < 0 , -1< x \leq 1$ - the graph concaves down.

$\frac {d^2y}{ dx^2} > 0 , x> 1$ - the graph concaves up.

Graph of functions (1) :

Given that curve is symmetrical about 𝑥- axis :