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idontgetconicsections
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?
@amistre64
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multiply by -1 u will have \( x=y^2+4 \) can u draw it?
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Or is it only in quadrant 1?
u should move it to right
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Well, the answer of a square root is nonnegative.. so definitely can't be 3 or 4
Like that?
So it's I and IV
Why is it wrong slaaiback?
I just told you. "Well, the answer of a square root is nonnegative.. so definitely can't be 3 or 4"
It's only I then?
Uhmm... why doesn't it look like a parabola?
because I freehanded the graph. Just to illustrate where it goes. You can make it parabola-like yourself
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?
sorry u must consider ristrictions for x ,y from \( y=\sqrt{x-4} \) u have y>=0 and x>=4
It's all of them. x^2 is nonnegative yes, but x^2 - 4 is not always nonnegative
This is really confusing.. which quadrants does x = y² + 1 occupy? I chose 1 because it's nonnegtaive, but 1 is not negative? So... ?
Oh wait, that's a square root, so it's quadrant 1 and 4??
omg, it's 1 and 4. The quadrants where x is positive
ye 1 and four looks correct to me.