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

##### 1 Answer

There are critical points at

#### 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

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

**For #y = 1#**

#(x - 1)^2 = 4#

#x - 1= +-2#

#x = 3 or -1#

**For #x = 1#**

#0 = y^2#

#y = 0#

The critical points are therefore

Hopefully this helps!