SOLUTION: I am not sure if i submitted it to the correct help section, but here it goes... determine whether each relation is a function. x^2 = 1 + y^2

Algebra ->  Functions -> SOLUTION: I am not sure if i submitted it to the correct help section, but here it goes... determine whether each relation is a function. x^2 = 1 + y^2      Log On


   

Question 58751This question is from textbook Elementary and Intermediate Algebra
: I am not sure if i submitted it to the correct help section, but here it goes...
determine whether each relation is a function.
x^2 = 1 + y^2 This question is from textbook Elementary and Intermediate Algebra

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
I am not sure if i submitted it to the correct help section, but here it goes...
determine whether each relation is a function.
x^2 = 1 + y^2
****Anytime you have an absolute value of y or y raised to an even power it is NOT a function****
Watch what happens if we try to solve for y:
x%5E2=1%2By%5E2
x%5E2-1=-1%2B1%2By%5E2
x%5E2-1=y%5E2
+/-sqrt%28x%5E2-1%29=sqrt%28y%5E2%29
+/-sqrt%28x%5E2-1%29=y The +/- means that there are two different possibilities for every x we put in, which is a no-no where functions are concerned.
Graphically you do not pass the vertical line test either:
graph%28300%2C200%2C-10%2C10%2C-10%2C10%2Csqrt%28x%5E2-1%29%2C-sqrt%28x%5E2-1%29%29
I just remember ****, the rest is for general information, depending on what kind of proof you have to give your teacher.
Happy Calculating!!!