Question

WILL GIVE BRAINLIEST PLEASE PLEASE PLEASE HELPPPPP

Answer 1

Complete the squares:

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

$y^2-4y=y^2-4y+4-4=(y-2)^2-4$

So we can rewrite the given expression as

$\bigg((x+1)^2-1\bigg)+\bigg((y-2)^2-4\bigg)+8=(x+1)^2+(y-2)^2+3$

Both $(x+1)^2$ and $(y-2)^2$ are non-negative for any choice of real $x,y$, which means the minimum value is 3 for $x=-1$ and $y=2$.