Which relation is not a function?
1)(x-2)^2+y^2=4
2) x^2+4y+y=4
3) x+y=4
4) xy=4
So, I could eliminate some answers. My answer is either its 1 or 2. Can anyone help me?
oops.. i meant:
x^2+4x+y=4
2)... 4y+y...?
1)y=(+ or -)sqrt(4-(x-2)^2) not a function
2)y=4-x^2-4x function
3)y=4-x function
4)y=4/x function
Thank you! :D
To determine if a relation is a function, we need to check if each input value (x) corresponds to exactly one output value (y). Let's analyze the given options:
1) (x-2)^2 + y^2 = 4
This is a circle centered at (2,0) with a radius of 2. Since one x-value can correspond to two different y-values (once on the top half and once on the bottom half of the circle), this relation is not a function.
2) x^2 + 4y + y = 4
Simplifying the equation, we have x^2 + 5y = 4. This equation represents a straight line. Each x-value corresponds to one specific y-value, so this relation is a function.
3) x + y = 4
This equation represents a straight line. Each x-value corresponds to one specific y-value, so this relation is a function.
4) xy = 4
Solving for y, we get y = 4/x. Each x-value corresponds to one specific y-value, so this relation is a function.
Therefore, the relation that is not a function is option 1) (x-2)^2 + y^2 = 4.