Question

Trace the curve y²=(x-a)(x-b)(x-c) with a, b, and c are all positive.  

Answer 1

We consider the following cases:

Case I : $a < b < c$

(1) It is symmetrical about the $x$ -axis.

(2) It meets the $x$ -axis in$(a, 0), (b,0)$ and $(c, 0).$

(3) When $x < a,$ $y^2$ is negative,

when $a<x<b, y^2>0$

when $b<x<c, y^2$ is negative,

when $x>c, y^2>0.$

Hence, there is no curve to the left-of $x =a$ and also between $x =b$ and $x=c.$

(4) If $x > c$ and increases then $y^2$ also increases.

Case II : $a = b < c$

$$y^2=(x-a)^2(x-c)$$

(1) It is symmetrical about the $x$ -axis.

(2) It meets the $x$ -axis in$(a, 0)$ and $(c, 0).$

(3) When $x < a,$ $y^2$ is negative,

when $a<x<c, y^2$ is negative.

$(a, 0)$ is isolated point.

(4) If $x > c$ and increases then $y^2$ also increases.

Case III : $a < b = c$

$$y^2=(x-a)(x-b)^2$$

(1) It is symmetrical about the $x$ -axis.

(2) It meets the $x$ -axis in$(a, 0)$ and $(b, 0).$

(3) When $x < a,$ $y^2$ is negative,

(4) If $x > b$ and increases then $y^2$ also increases.

Case IV : $a = b = c$

$$y^2=(x-a)^3$$

(1) It is symmetrical about the $x$ -axis.

(2) It meets the $x$ -axis in$(a, 0).$

(3) When $x < a,$ $y^2$ is negative,

(4) If $x >a$ and increases then $y^2$ also increases.