SOLUTION: Determine whether the relation is a function. x^2=1+y^2

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Question 78017: Determine whether the relation is a function.
x^2=1+y^2
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
In order to determine if this is a function lets solve for y.

x%5E2=1%2By%5E2

x%5E2-1=y%5E2 Subtract 1 from both sides

sqrt%28x%5E2-1%29=sqrt%28y%5E2%29 Take the square root of both sides
So we have

y=0%2B-sqrt%28x%5E2-1%29
Now lets graph y=0%2B-sqrt%28x%5E2-1%29. This can be broken up as
y=sqrt%28x%5E2-1%29 or y=-sqrt%28x%5E2-1%29

+graph%28+300%2C+200%2C+-6%2C+5%2C+-10%2C+10%2Csqrt%28x%5E2-1%29%2C+-sqrt%28x%5E2-1%29%29+ graph of y=sqrt%28x%5E2-1%29 and y=-sqrt%28x%5E2-1%29
Since the graph fails the vertical line test, it is not a function.