## idontgetconicsections Group Title Select all of the quadrants that the parabola whose equation is y=sqrt(x-4) occupies. I II III IV So the basic rules are x is always positive to the right so: x=ay^2 opens to the right x is always negative to the left; so x=-ay^2 opens to the left My problem is: I got rid of the radical by squaring, which gave me y^2 = x-4, then I subtracted x and y^2 from both sides to get -x=-y^2-4. Am I supposed to multiply all numbers by -1 to get rid of the negative numbers or do I just leave it like that? 2 years ago 2 years ago

1. idontgetconicsections Group Title

@amistre64

2. idontgetconicsections Group Title

|dw:1342621312995:dw|

3. mukushla Group Title

multiply by -1 u will have $$x=y^2+4$$ can u draw it?

4. idontgetconicsections Group Title

|dw:1342621412614:dw|

5. idontgetconicsections Group Title

Or is it only in quadrant 1?

6. mukushla Group Title

u should move it to right

7. idontgetconicsections Group Title

|dw:1342621507272:dw|

8. slaaibak Group Title

Well, the answer of a square root is nonnegative.. so definitely can't be 3 or 4

9. idontgetconicsections Group Title

Like that?

10. idontgetconicsections Group Title

Thanks

11. slaaibak Group Title

No it's not

12. idontgetconicsections Group Title

So it's I and IV

13. idontgetconicsections Group Title

Why?

14. idontgetconicsections Group Title

Why is it wrong slaaiback?

15. slaaibak Group Title

I just told you. "Well, the answer of a square root is nonnegative.. so definitely can't be 3 or 4"

16. idontgetconicsections Group Title

It's only I then?

17. slaaibak Group Title

|dw:1342621638765:dw|

18. slaaibak Group Title

Yes. Only 1

19. idontgetconicsections Group Title

Uhmm... why doesn't it look like a parabola?

20. slaaibak Group Title

because I freehanded the graph. Just to illustrate where it goes. You can make it parabola-like yourself

21. idontgetconicsections Group Title

Haha. Well, I have another question: Score: Select all of the quadrants that the parabola whose equation is y = x² - 4 occupies I chose all quadrants... is this correct or is it only 1 and 2 because it's nonnegative?

22. mukushla Group Title

sorry u must consider ristrictions for x ,y from $$y=\sqrt{x-4}$$ u have y>=0 and x>=4

23. slaaibak Group Title

It's all of them. x^2 is nonnegative yes, but x^2 - 4 is not always nonnegative

24. idontgetconicsections Group Title

This is really confusing.. which quadrants does x = y² + 1 occupy? I chose 1 because it's nonnegtaive, but 1 is not negative? So... ?

25. idontgetconicsections Group Title

Oh wait, that's a square root, so it's quadrant 1 and 4??

26. myko Group Title

omg, it's 1 and 4. The quadrants where x is positive

27. slaaibak Group Title

ye 1 and four looks correct to me.