The temperature of a point in the plane is given by the expression . What is the temperature of the coldest point in the pla

$$\\T=x^2+y^2-4x+2y\\\\ T+4+1=(x^2-4x+4)+(y^2+2y+1)\\\\ T+5=(x-2)^2+(y+1)^2\\\\ T=(x-2)^2+(y+1)^2\;-5\\\\$$

Now anything squared has to be  greater or equal to zero so T has to be greater to or equal to -5 degrees. 

This will occur when x=2 and y=-1

please help me!!!!

Alan's answer

http://web2.0calc.com/questions/the-temperature-of-a-point-nbsp-nbsp-in-the-plane-is-given-by-the-expression-nbsp-what-is-the-temperature-of-the-coldest-point-in-the-pla#r3

There is always more than one way to skin a cat 

Very nice, Melody........no Calculus required !!!!

Thanks Chris  :))