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 2 years ago
Select all of the quadrants that the parabola whose equation is y=sqrt(x4) 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 = x4, then I subtracted x and y^2 from both sides to get x=y^24. 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
Select all of the quadrants that the parabola whose equation is y=sqrt(x4) 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 = x4, then I subtracted x and y^2 from both sides to get x=y^24. Am I supposed to multiply all numbers by 1 to get rid of the negative numbers or do I just leave it like that?

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idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1@amistre64

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1342621312995:dw

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0multiply by 1 u will have \( x=y^2+4 \) can u draw it?

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1342621412614:dw

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1Or is it only in quadrant 1?

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0u should move it to right

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1342621507272:dw

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.0Well, the answer of a square root is nonnegative.. so definitely can't be 3 or 4

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1Like that?

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1So it's I and IV

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1Why is it wrong slaaiback?

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.0I just told you. "Well, the answer of a square root is nonnegative.. so definitely can't be 3 or 4"

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1It's only I then?

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1Uhmm... why doesn't it look like a parabola?

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.0because I freehanded the graph. Just to illustrate where it goes. You can make it parabolalike yourself

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1Haha. 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?

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0sorry u must consider ristrictions for x ,y from \( y=\sqrt{x4} \) u have y>=0 and x>=4

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.0It's all of them. x^2 is nonnegative yes, but x^2  4 is not always nonnegative

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1This is really confusing.. which quadrants does x = y² + 1 occupy? I chose 1 because it's nonnegtaive, but 1 is not negative? So... ?

idontgetconicsections
 2 years ago
Best ResponseYou've already chosen the best response.1Oh wait, that's a square root, so it's quadrant 1 and 4??

myko
 2 years ago
Best ResponseYou've already chosen the best response.0omg, it's 1 and 4. The quadrants where x is positive

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.0ye 1 and four looks correct to me.
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