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]

25 Feb 2006, 22:02

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

a) a<0

b) P>0

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25 Feb 2006, 22:47

I got C

If Y intersects x axis then we can say

0 = a(x-b)^2 + P
i.e (x-b)^2 = -P/a

there are two solutions to this equation
x = b+ SQRT(-P/a).............Eq1
OR
x = b-SQRT(-P/a)...............Eq2

For Y to intersect with x-axis x must be a real number.

St1: a<0. If P is -ve then both equations yield x = imaginary number. If P is +ve then both equations yield x = real number -----INSUFF

St2: P>0 If a is +ve then both equations yield x = imaginary number. If a is -ve then both equations yield x = real number -----INSUFF

Both statements together yield x = real number
Am I right laxie???
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26 Feb 2006, 02:03

Using both statements we can say that the parabola has the lowest point above the x-axis and is openedin the opposite direction of the x-axis. We can say that it has no intersections with the x-axis.

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Re: DS parabola [#permalink]

28 Feb 2006, 05:36

laxieqv wrote:

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

a) a<0

b) P>0

The problem is in fact: with the following statements, can we conclude whether the equation a(x-b)^2 + P = 0 has roots or not.

a(x-b)^2+ P = a( x^2-2xb+ b^2) +P = ax^2- 2abx + ab^2 + P

the determinant of this quadratic equation=
(-2ab)^2 - 4* a * (ab^2+P)= 4a^2*b^2 - 4a^2*b^2 - 4aP = -4aP

from either a) and b) we can't conclude if the determinant is -ve or +ve

Combine the two stmts, we have -4aP > 0
----> the equation has roots-----> y intersects with x-axis.

C it it.

Re: DS parabola [#permalink] 28 Feb 2006, 05:36

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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?

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

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