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# If a not equal to 0, and y = a (x+h)^2 + c , the graph of

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CEO
Joined: 15 Aug 2003
Posts: 3454
If a not equal to 0, and y = a (x+h)^2 + c , the graph of [#permalink]

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17 Dec 2003, 04:58
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If a not equal to 0, and y = a (x+h)^2 + c , the graph of the equation
intersects the X- axis at how many points?

a. h = 0

b. a < 0
Intern
Joined: 27 Nov 2003
Posts: 33
Location: Moscow

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17 Dec 2003, 05:15
cause we don't know anything about c:

a defines direction of parabola branches (up or down), c determines location of a NODE of the curve
CEO
Joined: 15 Aug 2003
Posts: 3454

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17 Dec 2003, 05:19
anvar wrote:
cause we don't know anything about c:

a defines direction of parabola branches (up or down), c determines location of a NODE of the curve

show some mathematical proof
Intern
Joined: 27 Nov 2003
Posts: 33
Location: Moscow

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17 Dec 2003, 05:29
Oh, well it is well known that if a<0 branches of parabola go down and if a>0 branches go up. At the same time if c<0, node of parabola is under X axis and if C>0 the Node is above X curve. In other words C defines a shift of parabola along Y curve.

=> even if a<0 it is possible that C<0 also. Therefore, branches of parabola can not cross X axis.
_________________

me

SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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12 Jan 2004, 14:14
y = a(x+h)^2 + c

for it to intersect with x axis y = 0
so
a(x+h)^2 + c = 0

from condition 1 ) h = 0
so x = sqrt( -c/a ) For this to be real number a < 0 but we dont know either a or c
so 1) is insufficient.
Same holds for 2)
Even if you combine condition 2) just makes x a real number but there infinie possibilities for the values of a and c

SO the answer should be E
Senior Manager
Joined: 11 Nov 2003
Posts: 356
Location: Illinois

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04 Feb 2004, 15:30
anandnk wrote:
y = a(x+h)^2 + c

for it to intersect with x axis y = 0
so
a(x+h)^2 + c = 0

from condition 1 ) h = 0
so x = sqrt( -c/a ) For this to be real number a < 0 but we dont know either a or c
so 1) is insufficient.
Same holds for 2)
Even if you combine condition 2) just makes x a real number but there infinie possibilities for the values of a and c

SO the answer should be E

Anand,

I did this problem exactly the same way you did. But I do not understand the last statement in your post (the bold text). I think, even if a < 0, that does not make x a real number. Because it is possible that c < 0. So we cannot answer the question even if after combining the statements. So this does not change the answer. Just wanted to confirm the reasoning.

Thanks
04 Feb 2004, 15:30
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# If a not equal to 0, and y = a (x+h)^2 + c , the graph of

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