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Intern
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ds- axis [#permalink] New post 13 Sep 2005, 17:32
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A
B
C
D
E

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
In the xy-plane, at what points does the graph of y =(x+a)(x+b) intersect the x- axis?
1. a+b =-1
2. The graph intersects y axis at (0,-6)
VP
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 [#permalink] New post 13 Sep 2005, 18:54
I will go with C.

y = (x + a) (x+b)
or
y = x^2 + (a+b)x + ab .....(a)

in order to cut x axis, y = 0. OR
x ^ 2 + (a + b) x + ab = 0

i) insufficient
ii) putting (0, -6) into equation (a) gives
ab = -6

taking both these,

cuts x axis when:
x^2 - x - 6 = 0
or x = 3, -2

As question asking "Points", having more than one anser is ok.
Senior Manager
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 [#permalink] New post 13 Sep 2005, 19:01
C
At one point the the graph cuts x-axis. Equation of x-axis is y=0
Substituting... we can simplify the function to
(X+a)(X+b)
or essentially what is (-a,-b) ?... or what are the values of a & b?
i) a+b = -1 Insuff
ii) Graph intersects Y axis - so x=0, y=-6 .. replace in the parent equation to get value of ab = -6 .. insuff

from i) & ii) we have 2 equations... solve for a & b... Together they are suff => C.
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 [#permalink] New post 14 Sep 2005, 07:11
It's a C.

The question asks for points.

1, insuff

b+a only gives us, y= sq(x) -1 + ab

2, insuff
value for ab = -6

1 and 2 gives y= sq(x) -x -6, set y=0 and solve the equation.
This will give you 2 roots, a positve and a neg x value.
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 [#permalink] New post 14 Sep 2005, 11:56
Yeah the OA is C.

You guys rock.
  [#permalink] 14 Sep 2005, 11:56
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