X2+4x+y2-2y=0 list the intercepts and test for symmetry?

Answer 1

y-intercepts:
2^2+4x=0
(x +4)( x)=0
x= -4 and x=0
intercepts: (0,0) (-4,0)

x-intercepts:
y^2-2y=0
(y-2)(y)=0
y=0 and y=2
intercepts: (0,0) (0,2)
Symmetry:

[Symmetry to the x-axis
(x, -y) is on the graph when (x, y) is. Test for symmetry
In the equation, leave x alone and replace y with -y. If you get the same equation, it is symmetric to the x-axis.
Symmetry to the y-axis
(-x, y) is on the graph when (x, y) is. Test for symmetry.
In the equation, replace x with -x. If the equation is the same, it is symmetric to tye y-axis.
Symmetry to the line y = x.
(y, x) is on the graph when (x, y) is. Test for symmetry.
Interchange x and y. If the equation is the same then it is symmetric to the line y = x.
Symmetry to the origin.
(-x, -y) is on the graph when (x, y) is. Test for symmetry.
Replace x with -x and y with -y. If the equation is the same, then it is symmetric to the origin.]

Answer 2

x^2 + 4x + y^2 - 2y = 0
(x^2 + 4x) + (y^2 - 2y) = 0
(x^2 + 4x + 4 - 4) + (y^2 - 2y + 1 - 1) = 0
((x + 2)^2 - 4) + ((y - 1)^2 - 1) = 0
(x + 2)^2 - 4 + (y - 1)^2 - 1 = 0
(x + 2)^2 + (y - 1)^2 - 5 = 0
(x + 2)^2 + (y - 1)^2 = 5

the Intercepts are (-4,0), (0,0), and (0,2)

There are many symmetries since the problem is a circle.

For a graph, go to www.quickmath.com

Answer 3

it is a circle with centre (-2,1).
and symetrical since a circle

Answer 4

do ur homework on ur own.