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Y=a(x-b)^2+P. Can you tell if y has intersects with x axis?

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Y=a(x-b)^2+P. Can you tell if y has intersects with x axis? [#permalink]  31 May 2009, 09:54
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Y=a(x-b)^2+P. Can you tell if y has intersects with x axis?
1). a<0
2). P>0

explain the method pls
Intern
Joined: 17 May 2009
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Location: USA
Schools: Kellogs, Wharton, Chicago
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Re: intersection [#permalink]  31 May 2009, 10:32
at x = b, y= P

If P < 0 then a should be greater than 0 to increase y and cut x-axis.

If P > 0 then a should be less than 0 to decrease y and cut x-axis.

so if both a < 0 and P > 0 Y will decrease and eventually meet x- axis.

Also,

y = 0 ==> x = (-p/a)^0.5 +b

x will have real values only when p/a <0 ==> either [P < 0 and a >0 ]or [P > 0 and a < 0]

IMO ans C.
Senior Manager
Joined: 16 Jan 2009
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Concentration: Technology, Marketing
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Re: intersection [#permalink]  31 May 2009, 14:01
traveller10 wrote:

y = 0 ==> x = (-p/a)^0.5 +b

x will have real values only when p/a <0 ==> either [P < 0 and a >0 ]or [P > 0 and a < 0]

IMO ans C.

Nice one.
IMO C

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Lahoosaher

Re: intersection   [#permalink] 31 May 2009, 14:01
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