Question

How do you identify the critical points for `(x-1)^2 = (y+1)^2`?

Answer 1

There are critical points at `(1, 0)`; `(3, 1)` and `(-1, 1)`.

Explanation:

Start by expanding.

`x^2 - 2x + 1 = y^2 + 2y + 1`

`x^2 - y^2 - 2x - 2y + 1 - 1 = 0`

`x^2 -y^2 - 2x - 2y = 0`

Now differentiate.

`2x - 2y(dy/dx) - 2 - 2(dy/dx)=0`

`-2y(dy/dx) -2(dy/dx) = 2 - 2x`

`dy/dx(-2y - 2) = 2 - 2x`

`dy/dx = (2 - 2x)/(2y - 2)`

`dy/dx = (2(1 - x))/(2(y - 1))`

`dy/dx = (1 - x)/(y - 1)`

Critical points will occur whenever the derivative equals `0` or is undefined. It will be undefined when `y = 1`. When the derivative equals `0`, `x = 1`.

Let's solve for the corresponding x-and y values.

For `y = 1`

`(x - 1)^2 = (1 + 1)^2`

`(x - 1)^2 = 4`

`x - 1= +-2`

`x = 3 or -1`

For `x = 1`

`(1 - 1)^2 = y^2`

`0 = y^2`

`y = 0`

The critical points are therefore `(1, 0)`, `(3, 1)` and `(-1, 1)`.

Hopefully this helps!