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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]

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New post Updated on: 01 Jun 2013, 03:28
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A
B
C
D
E

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Question Stats:

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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)

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-xy-plane-at-what-points-does-the-graph-of-y-88398.html

Originally posted by misterJJ2u on 21 Jan 2007, 00:36.
Last edited by Bunuel on 01 Jun 2013, 03:28, edited 2 times in total.
Edited the question and added the OA
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Re: graph question - gmat prep  [#permalink]

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New post Updated on: 22 Oct 2009, 13:47
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In the xy-plane, at what two points 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)

X-intercepts of the function \(f(x)\) or in our case the function (graph) \(y=(x+a)(x+b)\) is the value(s) of \(x\) for \(y=0\). So basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\).

\((x+a)(x+b)=0\) --> \(x^2+bx+ax+ab=0\) --> \(x^2+(a+b)x+ab=0\).

Statement (1) gives the value of \(a+b\), but we don't know the value of \(ab\) to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of \(y\) when \(x=0\) --> \(y=(x+a)(x+b)=(0+a)(0+b)=ab=-6\). We know the value of \(ab\) but we don't know the value of \(a+b\) to solve the equation.

Together we know the values of both \(a+b\) and \(ab\), hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

Answer: C.

Hope it's clear.
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Originally posted by Bunuel on 22 Oct 2009, 13:28.
Last edited by Bunuel on 22 Oct 2009, 13:47, edited 1 time in total.
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New post 21 Jan 2007, 02:49
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For me (C) :)

y= (x+a)(x+b) intersect the x-axis is equivalent to search the solutions of :
(x+a)(x+b) =0
<=> x^2 + (a+b)*x + a*b = 0 (1)

From 1
a+b = -1
implies:

(1) <=> x^2 -x + a*(-a-1) = 0

we need to know a.

INSUFF.

From 2
(0, -6) is the Y interceptor. That implies:

a*b = -6.

we need another equation with a & b, in other words, the values of a and b.

INSUFF.

Both (1) and (2)
a+b = -1
a*b = -6

These equations give us (a,b), what we need to find the roots.

SUFF.
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Re: DS: Coordinate Geometry (intersection)  [#permalink]

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New post 19 Aug 2008, 08:58
2
lordw 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).

Guys I felt lost in this problem with all the letters. Any idea of how to put this in a more "visible" way? For instance, may be picking numbers? Many tks. Lw.


Graph intersects x-axis when y=0 , therefore two points where graph intersects is x=-a, x=-b

Statement 1) is insuff

Statement 2) just tells one point

Combining both , points are 6 and -5

Answer is C)
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Re: DS: Coordinate Geometry (intersection)  [#permalink]

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New post 19 Aug 2008, 10:00
3
1
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).

for point intersect the x axis, we need to find x when y = 0
so 0 = (x+a) (x+b)
x = -a or -b
So, we need to find a and b

(1) +(2) is sufficient to find two unknown

(C)
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Re: xy plane  [#permalink]

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New post 11 Oct 2008, 12:15
3
The answer is (E)

y=(x+a)(x+b)

y=x^2+ax+ab+ab

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

statement 1

a+b = -1

y=x^2-x+ab

0=x^2-x+ab

insufficient

statement 2

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

ab=-6

0=x^2+x(a+b)-6

insufficient

both statements

0=x^2-x-6

6=x(x-1)

x can be -2 or 3

insufficient

:)
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Re: xy plane  [#permalink]

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New post 11 Oct 2008, 14:36
I agree with fresinha12, I get 3 and -2

in the question, it asks for points, so I guess it is okay to say that the answer is C
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Re: xy plane  [#permalink]

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New post 12 Oct 2008, 11:51
1
redbeanaddict wrote:
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 the y axis at (0, -6)

the graph intersects x axis hen y=0 => x=-a and x=-b are the points

to get the points we must know a and b but how to find

1) does not help
2)does not help => gives ab=-6

combine 1 and 2 we get a and b hence is SUFFI to answr
IMO C
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Re: GMAT PREP_DS_XY plane  [#permalink]

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New post 20 Aug 2009, 19:06
The answer provided above is incorrect.

The question does not ask for the values of a and b.

It asks for the values of x when y=0. In other words, it asks for the roots of the quadratic.

Statement 2 is sufficient, as it says that ab=-6.

Thus: x^2-6x-6=0 -> Roots are -0.87 and 6.87.

Answer is B.
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graph question - gmat prep  [#permalink]

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New post 22 Oct 2009, 13:14
1
Hi,

so here's what I did:

y= x^2+ xb+xa+ab

I replace y with -6 and x with 0

-6=0+0+ab
so ab= -6

I replace ab with -6 and y with 0 in the first equation

0= x^2+ xb+xa+(-6)
6=x(x+b+a)

from statement A, we know that a+b=1
so 6 = x(x+1)
x^2+x-6 = 0

(x-3)(x+2)
x=3 or x = -2

I then got stuck :?
so what is the value of x?
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Re: graph question - gmat prep  [#permalink]

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New post 22 Oct 2009, 13:45
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Bunuel wrote:
In the xy-plane, at what two points 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)

Q is in what two points graph intersect the x-axis. Which means we should be able to find two points: (x1,0) and (x2,0), so basiacally the roots of equation: (x+a)(x+b)=0.

y=(x+a)(x+b)=x^2+x(a+b)+ab. --> We must now the values of (a+b) and ab.

(1) a+b=-1 --> not sufficient.
(2) x=0, y=-6 --> -6=ab not sufficient

(1)+(2) a+b=-1, ab=-6 --> y=x^2+x(a+b)+ab=x^2-x-6 --> x intersection means y=0 --> 0=x^2-x-6 --> (x+2)(x-3)=0 -
-> x1=-2, x2=3. Sufficient

Answer C.


OMG!!

I didnt realize that I need to find two points!!
thats why when i ended up with -2 and 3, i got confused because i was left with two values!!!

how did you find ab=-6?
(2) x=0, y=-6 --> -6=ab not sufficient

did you use the formula
y=mx+b?

thanks,
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Re: graph question - gmat prep  [#permalink]

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New post 22 Oct 2009, 13:51
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Re: graph question - gmat prep  [#permalink]

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New post 23 Oct 2009, 04:34
Great Bunuel! I have been seeing your post and I have to say I am quite amazed at your math skills! Have you taken the GMAT yet? I would very much like to know your score.
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Re: graph question - gmat prep  [#permalink]

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New post 23 Oct 2009, 10:47
Yup Bunuel is awesome as always :)

Here, we need to find the solution for (x+a)(x+b) = 0.
Solutions are x=-a, x=-b.
Thus we need to find unique values for a and b to answer the question.

1) we get a relation between a and b but not sufficient to derive unique values of a and b. insuff.

2) we get a point on the graph, so substituting values of x and y in the graph we get ab = -6. Again, this is insufficient to find unique values of a and b. insuff.

Combining, we have, a+b=-1 and ab= -6, two variables, two equations. So answer is C. The points will be (-a,0), (-b,0).
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Re: GMAT PREP_DS_XY plane  [#permalink]

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New post 23 Nov 2009, 19:31
powerka wrote:
The answer provided above is incorrect.

The question does not ask for the values of a and b.

It asks for the values of x when y=0. In other words, it asks for the roots of the quadratic.

Statement 2 is sufficient, as it says that ab=-6.

Thus: x^2-6x-6=0 -> Roots are -0.87 and 6.87.

Answer is B.


Fig is correct. You made a mistake on the equation, it is x^2-(a+b)x+ab.

Essentially, you are looking for a and b itself, or equations that tell you a+b and ab.

Hence, C
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Re: graph question - gmat prep  [#permalink]

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New post 25 Dec 2010, 04:49
Bunuel wrote:
In the xy-plane, at what two points 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)

X-intercepts of the function \(f(x)\) or in our case the function (graph) \(y=(x+a)(x+b)\) is the value(s) of \(x\) for \(y=0\). So basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\).

\((x+a)(x+b)=0\) --> \(x^2+bx+ax+ab=0\) --> \(x^2+(a+b)x+ab=0\).

Statement (1) gives the value of \(a+b\), but we don't know the value of \(ab\) to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of \(y\) when \(x=0\) --> \(y=(x+a)(x+b)=(0+a)(0+b)=ab=-6\). We know the value of \(ab\) but we don't know the value of \(a+b\) to solve the equation.

Together we know the values of both \(a+b\) and \(ab\), hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

Answer: C.

Hope it's clear.



I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer.Can you please help on this.
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Re: graph question - gmat prep  [#permalink]

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New post 25 Dec 2010, 08:49
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gautamsubrahmanyam wrote:
Bunuel wrote:
In the xy-plane, at what two points 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)

X-intercepts of the function \(f(x)\) or in our case the function (graph) \(y=(x+a)(x+b)\) is the value(s) of \(x\) for \(y=0\). So basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\).

\((x+a)(x+b)=0\) --> \(x^2+bx+ax+ab=0\) --> \(x^2+(a+b)x+ab=0\).

Statement (1) gives the value of \(a+b\), but we don't know the value of \(ab\) to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of \(y\) when \(x=0\) --> \(y=(x+a)(x+b)=(0+a)(0+b)=ab=-6\). We know the value of \(ab\) but we don't know the value of \(a+b\) to solve the equation.

Together we know the values of both \(a+b\) and \(ab\), hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

Answer: C.

Hope it's clear.



I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer. Can you please help on this.


The question asks: at what two points the graph of \(y=(x+a)(x+b)\) intersect the x-axis?

So we should find two points: (x1, 0) and (x2, 0), (points of intersection of the given graph with X-axis). Basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\) --> \(x^2+(a+b)x+ab=0\).

When we combine statement we have: \(a+b=-1\) and \(ab=-6\), so you should solve quadratic equation \(x^2-x-6=0\) --> \(x_1=-2\) and \(x_2=3\) --> points of intersection are (-2, 0) and (3, 0). Check on the diagram:
Attachment:
MSP272919df1bed26600c1g0000673adafe5fce1fc2.gif
MSP272919df1bed26600c1g0000673adafe5fce1fc2.gif [ 3.61 KiB | Viewed 123077 times ]
Hope it's clear.
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Re: xy plane  [#permalink]

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New post 17 Mar 2011, 22:27
This problem is easy when you don't start by rapidly trying to solve it.
Think about it before you put pen to paper...
"what two points does the graph intersect the x axis..."
This is a simple problem asking for 2 values of x...
statement 1 gives us a+b=-1
statement 2 gives us ab

Start with the stem...multiply x+a and x+b, it will become very clear.
Y = X^2 + X (a+b) + ab
statement 1 gives you a+b, we need ab
statement 2 says graph intersects y at (0,-6). Plug this into the original statement, you will get ab.
Solved = C!
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Re: xy plane   [#permalink] 17 Mar 2011, 22:27
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