Question 693642
X =(y+2)^2 
Solve and state the domain and range, and write in interval notation. Is this equation a function?
This is a parabola that opens rightwards with the vertex at (0,0)
Its standard form of equation: (y-k)^2=4p(x-h)X 
(y+2)^2=x
This is a sqrt function where the radican x≥0
domain: [0,∞)
range: (-∞,∞)
equation is not a function.
To be a function, for each x-value, there can be only one y-value, and given equation has 2 different y-values for each x.