SOLUTION: y^2=x Which I figure to be y=sqrt(x)... When I graph it on my TI-89, it does not create a continuous line but stops 0. My answer key shows it continuing through the negative

Algebra ->  Linear-equations -> SOLUTION: y^2=x Which I figure to be y=sqrt(x)... When I graph it on my TI-89, it does not create a continuous line but stops 0. My answer key shows it continuing through the negative       Log On


   

Question 254973: y^2=x
Which I figure to be y=sqrt(x)...
When I graph it on my TI-89, it does not create a continuous line but stops 0. My answer key shows it continuing through the negative spectrum of y. Is the answer key wrong, or am I wrong in making the equations y=sqrt(x).
Sorry for the pathetically dumb questions. I haven't done algebra in over a decade, and the book I bought is very sparse on explanation. It's more of a workbook than anything.
Found 3 solutions by Alan3354, richwmiller, drk:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
If you use y= sqrt(x) you have to also use y = -sqrt(x)

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
You left out other half.
It should be plus or minus sqrt(x)
y=+\-sqrt(x)
It is better to leave it as y^2=x for graphing
Don't feel bad after 10 years. I am refreshing my memory after 45 years of not doing math

Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
The answer key is incorrect since we cannot take negative sqrt over the real axes.
+graph%28+300%2C+200%2C+-5%2C+5%2C+-5%2C+5%2C+sqrt%28x%29%29+