x^2+y ^2+ 4x + 6y - 12 = 0 how do i find the domain?
y^2 + 6 y = -x^2 -4x + 12
y^2 + 6 y + 9 = -x^2 -4x +21
-(y+3)^2 = + (x^2 + 4 x) - 21
- (y+3)^2 + 4 = x^2 + 4x + 4 -21
- (y+3)^2 + 4 = (x+2)^2 -21
(x+2)^2 + (y+3)^2 = 25
circle center (-2,-3) radius 5
domain from -7 to +3
thanks that helped so much
To find the domain of the equation x^2 + y^2 + 4x + 6y - 12 = 0, we need to determine the values that x and y can take.
To begin, let's focus on the x-terms. The equation contains an x^2 term and a 4x term. Since both terms involve x raised to a power, it means that x can take any real value. Therefore, the domain for x is (-∞, ∞), or all real numbers.
Next, let's look at the y-terms. The equation contains a y^2 term and a 6y term. Similar to x, both terms involve y raised to a power, meaning that y can also take any real value. Consequently, the domain for y is also (-∞, ∞), or all real numbers.
Therefore, the domain for this equation is (-∞, ∞) for both x and y.