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
#(x - 1)^2 = 4#
#x - 1= +-2#
#x = 3 or -1#
For
#0 = y^2#
#y = 0#
The critical points are therefore
Hopefully this helps!