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In the xy-plane, at what two points does the graph of

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In the xy-plane, at what two points does the graph of [#permalink] New post 04 Jul 2006, 03:03
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In the xy-plane, at what two points does the graph of y=(x+a)(x+b) intersect the x-axis?

1) a+b=-1
2) The graph intersects the y-axis at (0;6).

Please explain. Thank you.
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Re: DS: intersect of x-axis. [#permalink] New post 04 Jul 2006, 08:29
M8 wrote:
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) intersect the x-axis?

1) a+b=-1
2) The graph intersects the y-axis at (0;6).

Please explain. Thank you.


st. 1
Insuff, because infinite number of possibilities..

st. 2
y=x^2+(a+b)*x+a*b
if y-intersects at (0,6), then a*b = 6.
so equation is not --> y=x^2+(a+b)*x+6
before I try to combine st. 1, I know that the parabola is going to open up. I feel that this is Insuff.

combine both:
I get y=x^2-1*x+6, but this does not have integer roots for X...
so by the combination of both, it does not intersect the x-axis.

C is my answer... but would E work here?
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 [#permalink] New post 04 Jul 2006, 10:25
0 = (x+a)(x+b)

1. insufficient -- too many possibilities
2. can add new information to formula and make it:
0=(x-6)(x+b); since we want to get (x+b) = 0 there are still infinite possibilities.

However, when add information know from (1), can determine value of B. Since we know value of a from (2), we can solve the equation. C.
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 [#permalink] New post 04 Jul 2006, 10:34
amartin6165 wrote:
0 = (x+a)(x+b)

1. insufficient -- too many possibilities
2. can add new information to formula and make it:
0=(x-6)(x+b); since we want to get (x+b) = 0 there are still infinite possibilities.

However, when add information know from (1), can determine value of B. Since we know value of a from (2), we can solve the equation. C.


why did you assign a=-6?
if you don't mind, can you tell me what is your value of B?
also are you saying that the function y=(x+a)*(x+b) will have
x-intercepts once you apply both statements?

Thanks
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Re: DS: intersect of x-axis. [#permalink] New post 04 Jul 2006, 12:52
M8 wrote:
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) intersect the x-axis?

1) a+b=-1
2) The graph intersects the y-axis at (0;6).

Please explain. Thank you.


a + b =-1 (infinite possibilities)

ii - ab = 6 (infinite possibilities)

i and ii gives unique answers

So C
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 [#permalink] New post 04 Jul 2006, 13:08
(C)

I. Cannot calculate a,b given a+b = -1
II. ab = 6 ..Insufficient.


Combining both, we can get values for a, b.
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 [#permalink] New post 04 Jul 2006, 18:06
C

Graph will intersect x axis at -a and -b. Basically we need to find the value of a and b.

St1: a+b = -1 Can not find values of a and b: INSUFF

St2: 6 = ab Can not find the values of a and b: INSUFF

Combined:
Two distinct equations linear equations of two variables can be solved to get the values of variables.
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 [#permalink] New post 05 Jul 2006, 00:25
y=(x+a)(x+b)
y=x^2+x(a+b) +ab

1) a+b is given

Infinte posiibilites hence ab can be anything
2) ab = 6

a+b can be anything.
Not suff

Together

we get a definite value of y for all x hence sufficent
  [#permalink] 05 Jul 2006, 00:25
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