Hello, Honorable24!
I completed the square and got: . .??
Find: .
We have: .
. . . .
. . . . . . . .
. . . . . . . . .
This is a circle: .
There are no -intercepts.
There are no -intercepts.
A circle does not have vertices.
4x^2 + 4y^2 +20x -16y + 37 =0
I completed the square and got:
(2x+10)^2 + (2y-8)^2 =127
x int ?
y int ?
Vertices ?
Center (-5, 4)
Domain ?
Range ?
I tried making y=0 to find the x intercepts but didn't work out. Same thing for y intercepts
Completely lost....
I recommend you go back and check the problem. I suspect it should be .
If really is the problem, -6x+ 4x= -2x so this is the same as . The fact that there is no "y to the first power" term is not a problem. It only means that y is already a "perfect square"" . Complete the square in x.