graphing equation?

Question

how would i graph an equation with a y in it
the equation is x^2+6x+y^2+4Y-3=0
It should be a circle it says
thanks i have forgotten how to do this

can u explain more i dont understand i am trying to do what u say but i cant get the answer tell me if i am doing it right for x=2 i got the answer -4+/-sqrt92/2 is that right i dont get how u get the answer when i do the other numbers x=345 etc i still get a +/- which gives me two solutions i have a graphing calculator please explain more thanks

Answer 1

First you need to separate the y's onto the left and the x's onto the right of the equation...

y^2+ 4y = x^2 + 6x + 3
now pick a value of x and substitute in the right side and get a value:

x = 0:
y^2 + 4y = 0^2 +6(0) + 3 = 3
Solve for y:
y^2 + 4y - 3 = 0 (use Quadratic equation or plot on a graphing calculator and find the zeros.. should be 2 of them)

x = 1:
y^2 + 4y = 1 + 6 + 3 = 10
again, use quadratic equation to get the roots (values of y that will solve the equation, should be two)...

at some value of x you will get just ONE point, this is the furthest point left and the point furthest right on your graphed figure...

at some value of x you will find it impossible to get a value of y because it is to the left or right of your graphed figure.

Answer 2

dude it truly is extremely common. u divide both aspect by 4 and also you get y=3. considering there is no x fee (or u can say that the slope of the graph is 0, considering y=3 is that very same as y=0x+3), the graph will be a horizontal line, parallel to the x-axis yet 3 gadgets above the x-axis. if that continues to be uncertain then merely message me lower back.