SOLUTION: How do I find the relation of y to x in the function of x=(y+2)^2. Would this be a continuous function? i know that in this kind of equation were I to show it on a graph, x would

Algebra ->  Functions -> SOLUTION: How do I find the relation of y to x in the function of x=(y+2)^2. Would this be a continuous function? i know that in this kind of equation were I to show it on a graph, x would       Log On


   

Question 661157: How do I find the relation of y to x in the function of x=(y+2)^2.
Would this be a continuous function? i know that in this kind of equation were I to show it on a graph, x would be to plotted 2 units to the left. But what would my y coordinants be?
Found 2 solutions by swincher4391, MathLover1:
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
Right now you have a relation mapping x to y. You want to find y to x, so solve for y.

x+=+%28y%2B2%29%5E2
Take the square root.
sqrt%28x%29+=+%28y%2B2%29
y+=+-2+%2B-sqrt%28x%29
This function is continuous on (0,oo) since anything outside of that domain would result in the square root of a negative number.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
If the graph is a continuous, unbroken line, then we can to say that f%28x%29 is a continuous function.

+x=%28y%2B2%29%5E2..take a square of both sides
+sqrt%28x%29=y%2B2
+y=sqrt%28x%29-2
domain: all real non-negative numbers x where x%3E=0 (+sqrt%28x%29=0 if x=0)
range: all real numbers y where y%3E=-2 (+y%2B2=0 if y=-2)

+graph%28+600%2C+600%2C+-5%2C+30%2C+-5%2C+5%2Csqrt%28x%29-2%29%29+
the graph starts at y=-2 continues as unbroken line, then we can to say that +y=sqrt%28x%29-2 is a continuous function