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

1 Answer
Feb 27, 2017

Answer:

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!