In the xy plane, at what points does the graph of y= : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 16 Jan 2017, 04:35

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In the xy plane, at what points does the graph of y=

Author Message
TAGS:

### Hide Tags

Intern
Joined: 26 Nov 2009
Posts: 14
Followers: 1

Kudos [?]: 70 [1] , given: 0

In the xy plane, at what points does the graph of y= [#permalink]

### Show Tags

24 Dec 2009, 09:00
1
KUDOS
11
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

70% (02:11) correct 30% (01:22) wrong based on 587 sessions

### HideShow timer Statistics

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)
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Feb 2013, 04:23, edited 2 times in total.
Edited the question and added the OA
Math Expert
Joined: 02 Sep 2009
Posts: 36508
Followers: 7063

Kudos [?]: 92859 [2] , given: 10528

Re: GMAT Prep Question, Any Shortcuts? [#permalink]

### Show Tags

24 Dec 2009, 09:16
2
KUDOS
Expert's post
12
This post was
BOOKMARKED
JimmyWorld wrote:
This question took me much longer then 2 minutes to figure out and sucked up a lot of time for me. Are there any shortcuts that would make this problem easier to solve?

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)

OA
[Reveal] Spoiler:
C

Thanks.

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.

For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature).

Hope it helps.
_________________
Manager
Joined: 17 Dec 2009
Posts: 55
Followers: 0

Kudos [?]: 17 [0], given: 4

Re: GMAT Prep Question, Any Shortcuts? [#permalink]

### Show Tags

24 Dec 2009, 09:37
y=(x+a)(x+b) when y=0
To solve this one, what do we need to know?
Obviously a or b, which are not stated in the information (1) & (2)
So one rule advise by MGMAT Book, always expand when the information given is factorized or Factorized when the information given is expended.
We know from a quadratic expression, the x axis intersect when y=0
So let's expand, and we have x^2 + (a + b)x + ab = 0
So now, we need to know ab and (a+b) to solve this equation
Therefore the correct answer is C, since only both information taken together permit to answer the question.
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 956 [0], given: 25

Re: DS question : need help [#permalink]

### Show Tags

15 Oct 2010, 00:05
satishreddy wrote:
need help on DS question

Right, so when the graph intersects the x-axis, y is 0. Those points are clearly x=-a and x=-b

(1) a+b=1. Not sufficient to know a or b
(2) We have point (0,-6). Plus it into the equation to get -6=ab. Again not sufficient to know either number

(1+2). Now we have 2 equations, sufficient. a-6/a=1; a^2-a-6=0; a=3 or a=-2 Hence b=-2 or b=3 .. Either way the points are 3 and -2

_________________
Manager
Joined: 06 Aug 2010
Posts: 225
Location: Boston
Followers: 3

Kudos [?]: 183 [0], given: 5

Re: GMAT Prep Question, Any Shortcuts? [#permalink]

### Show Tags

02 Nov 2010, 06:47
JimmyWorld wrote:
This question took me much longer then 2 minutes to figure out and sucked up a lot of time for me. Are there any shortcuts that would make this problem easier to solve?

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)

OA
[Reveal] Spoiler:
C

Thanks.

We know that the graph will intersect the x-axis when y=0, which occurs when x = -a and x = -b.

1) This doesn't tell us what a and b are. For example, a could be 5 and b could be -6, or a could be -4 and b could be 3. Insufficient.

2) This tells us that either a or b is 6, but we don't know the other point. Insufficient.

Together, we know that one of the values is 6, and that a + b = -1. So we can get the other value. Sufficient.
Manager
Joined: 06 Apr 2010
Posts: 144
Followers: 3

Kudos [?]: 654 [0], given: 15

### Show Tags

03 May 2011, 05:50
1
This post was
BOOKMARKED
Attachment:

M-Q19.JPG [ 19.12 KiB | Viewed 8795 times ]
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1353
Followers: 17

Kudos [?]: 240 [0], given: 10

### Show Tags

03 May 2011, 09:19
y = x^2 + (a+b)x + ab is a parabola.Intersecting x axis at two points.
hence we need to find the roots of this equation.

a. a+b = -1 means a= -1 | b = -1 | a=b= -0.5 each etc. Hence not sufficient.

b y = -6 for x = 0

means ab = -6 (substituting in the original quadratic equation.

a= -3,b=1 | a = -6, b = 1 etc. hence not sufficient.

together 1+2

y = x^2 -x -6 = (x-3) * (x + 2) hence two points for X axis.

thus C.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Senior Manager
Joined: 24 Mar 2011
Posts: 457
Location: Texas
Followers: 5

Kudos [?]: 164 [0], given: 20

### Show Tags

03 May 2011, 09:21
1
This post was
BOOKMARKED
y=(x+a)(x+b) = x^2+(a+b)x+ab
this is parabola equation. intersects x-axis means y=0

st 1--> a+b = -1
x^2-x+ab=0 cannot solve the equation; not sufficient

st 2--> graph intersects y-axis at -6
i.e. ab = -6
x^2+(a+b)-6=0 cannot solve the eq, not sufficient.

Both together, x^2-x-6=0
Solving gives x = -2 or x=3

Hence C.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7118
Location: Pune, India
Followers: 2128

Kudos [?]: 13618 [0], given: 222

### Show Tags

03 May 2011, 17:33
Expert's post
2
This post was
BOOKMARKED
udaymathapati wrote:
Attachment:
M-Q19.JPG

At what two points does the graph of y = (x+a)(x+b) intersect the x axis?

You don't need to worry what the equation represents. Just think, what does 'intersection with x axis' imply? It means the y co-ordinate is 0.

0 = (x+a)(x+b)
or x = -a or -b
Hence the graph must intersect the x axis at points (-a, 0) and (-b, 0). We need the values of a and b now.

Statement 1: a + b = -1
Two variables, only one equation. Not sufficient.

Statement 2: Graph intersects the y axis at (0, -6).
At y axis, x = 0. This means when x = 0, y co-ordinate is -6.
Put these values in y = (x+a)(x+b) to get -6 = ab.
Again, two variables, one equation. Not sufficient alone.

Using both statements, we have two variables and two different equations so we will be able to find the values of a and b. It doesn't matter which is 'a' and which is 'b'. We find that the two of them are -3 and 2. Since we need the points (-a, 0) and (-b, 0), the required points are (3, 0) and (-2, 0). Sufficient.

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Intern
Joined: 19 Jul 2011
Posts: 24
Followers: 0

Kudos [?]: 5 [0], given: 2

In xy plane, at what 2 points does the graph of y=(x+a)(x+b) [#permalink]

### Show Tags

24 Jul 2011, 12:06
1
This post was
BOOKMARKED
In xy plane, at what 2 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)

[Reveal] Spoiler:

Last edited by fluke on 24 Jul 2011, 15:37, edited 1 time in total.
Hint Hidden
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
Followers: 161

Kudos [?]: 1701 [1] , given: 376

### Show Tags

24 Jul 2011, 15:37
1
KUDOS
tt2011 wrote:
In xy plane, at what 2 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)

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

Question is asking for the solution of x
(x+a)(x+b)=0
x^2+ax+bx+ab=0
x^2+(a+b)x+ab=0

1) a+b=-1
We got a+b. Yet, we don't know about "ab"
Not Sufficient.

2) Intersect y at (0,-6)
x=0; y=-6
-6=(0+a)(0+b)
ab=-6
We got ab. We don't know a+b.

Together;
Sufficient.

Ans: "C"
_________________
Intern
Joined: 15 Mar 2010
Posts: 10
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

24 Jul 2011, 20:22
We need to find the values of x when x = a or when x =b

Stmt 1: a+b = -1 gives numerous values of a and b . Insuff

Stmt 2: plugging in (0, -6) in the equation:

ab= -6 still cannot determine values of x

Combining 1 and 2 gives values of x as a quadratic eqn forms:

x^2-x-6=0. Hence C
Intern
Joined: 19 Jul 2011
Posts: 24
Followers: 0

Kudos [?]: 5 [0], given: 2

### Show Tags

24 Jul 2011, 23:33
thanks fluke...i arrived at x=-a or x= -b knowing we need to find the value of x..then tried plugging in numbers..your approach is better...
Manager
Joined: 20 Aug 2011
Posts: 146
Followers: 3

Kudos [?]: 95 [2] , given: 0

### Show Tags

26 Oct 2011, 02:04
2
KUDOS
1
This post was
BOOKMARKED
kotela wrote:
In the XY plane 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)

can anyone plz explain?

Intersect the x axis means y is 0.
the x coordinates will be -a and -b respectively.

1. a+b=-1
Insufficient

2.
-6=(0+a)(0+b)
ab=-6
Insufficient.

1+2
a+b=-1
ab=-6
a=-3, b=2
a=2, b=-3
Sufficient as we want two points we can say (-3,0) and (-2,0) without knowing the exact values of a & b.
_________________

Hit kudos if my post helps you.
You may send me a PM if you have any doubts about my solution or GMAT problems in general.

Last edited by blink005 on 26 Oct 2011, 06:17, edited 2 times in total.
Intern
Joined: 11 Jan 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Intersection at X axis? [#permalink]

### Show Tags

16 Feb 2012, 17:41
Here is my attempt:

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).

Essentially they are asking us to solve for x when y=0:

y = (x+a)(x+b) -> x^2 + xb + xa + ab = 0 (What is X?)

1) a = -b - 1 -> This is insufficient because it still leaves us with one unknown: b

2) -6 = ab -> This allows us to solve for ab in the line equation, but we still don't know what a and b are individually, which we need because we have an xb and xa term.

1) + 2) -> Knowing a = -6/b, we can plug into 1) and find b and a separately. Then the only unknown is x which we can then solve for.

Intern
Joined: 11 Jan 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Intersection at X axis? [#permalink]

### Show Tags

16 Feb 2012, 17:58
The graph intersects at 0,-6. That is when x = 0, and y=-6. X being zero conveniently cancels out all the terms in our original equation:

x^2 + xb + xa + ab = y

and we are left with ab = -6
Senior Manager
Joined: 08 Jun 2010
Posts: 397
Location: United States
Concentration: General Management, Finance
GMAT 1: 680 Q50 V32
Followers: 3

Kudos [?]: 88 [0], given: 13

Re: In the xy plane, at what points does the graph of y= [#permalink]

### Show Tags

26 Feb 2012, 22:50
I tried plugging in values in this one.

Started from option 2:
ab = -6
Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.

Option 1:
substitute for x and y in equation we get
a+b = -1
Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.

Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C.
Math Expert
Joined: 02 Sep 2009
Posts: 36508
Followers: 7063

Kudos [?]: 92859 [0], given: 10528

Re: In the xy plane, at what points does the graph of y= [#permalink]

### Show Tags

26 Feb 2012, 23:03
Expert's post
1
This post was
BOOKMARKED
mourinhogmat1 wrote:
I tried plugging in values in this one.

Started from option 2:
ab = -6
Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.

Option 1:
substitute for x and y in equation we get
a+b = -1
Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.

Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C.

This is not a good idea to plug the numbers for this problem, it's better to understand the concept and you won't need any math at all (certainly you won't need to solve a+b=-1 and ab=-6).

Next, there are infinitely many values of a and b possible to satisfy ab=-6, not just four: notice that we are not told that a and b are integers only, so for example a=1/2 and b=-12 is also a solution.
_________________
Intern
Joined: 22 Jan 2012
Posts: 35
Followers: 0

Kudos [?]: 20 [1] , given: 0

Re: In the xy plane, at what points does the graph of y= [#permalink]

### Show Tags

26 Apr 2012, 20:33
1
KUDOS
y= x^2+ax+bx+ab
y= x^2+x(a+b)+ab

the question asks us to find the value of two points where the graph intersects x axis. So y=0

x^2+x(a+b)+ab=0 =?

1) a+b=-1 , but we dot have any value of ab, hence not sufficient

2) putting the value x=0 , y=-6
y= x^2+x(a+b)+ab
ab=-6
but we dot have any value of a+b , hence not sufficient

1&2) x^2+x(a+b)+ab=0
now we have a+b=-1
ab =-6
hence sufficient to find 2 values of x.

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 3693
Followers: 1286

Kudos [?]: 5827 [0], given: 66

### Show Tags

30 Apr 2012, 11:17
Stiv 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)

So, it's important to know --- when the quadratic is given in factored form --- y= (x + a) (x + b) --- then we know the two roots, x = -a, and x = -b. Roots are the x-intercepts, the places where the graph intersects the x-axis. Basically, the prompt is asking us to find the values of a & b.

Statement #1: a + b = -1

One equation for two unknowns. Not enough to solve. Not sufficient.

Statement #2: The graph intersects the y-axis at (0, -6)

Plugging in x = 0 (the condition of the y-axis), we get y = (0+a)(0+b) = ab = -6

Again, one equation for two unknowns. Not enough to solve. Not sufficient.

Combined statements:
a + b = -1
ab = -6

Two equations with two unknowns ---> we can solve for the values of a & b, which will answer the question. Sufficient.

Here's another practice question on quadratics for practice.
http://gmat.magoosh.com/questions/120
When you submit your answer to that question, the next page will have a full video explanation.

Let me know if you have any further questions.

Mike
_________________

Mike McGarry
Magoosh Test Prep

Re: Question   [#permalink] 30 Apr 2012, 11:17

Go to page    1   2    Next  [ 36 posts ]

Similar topics Replies Last post
Similar
Topics:
1 In the xy-plane, at what two points does the graph 1 17 Jul 2012, 06:56
1 In the xy plane, at what points does the graph of y=(x+a)(x+ 2 25 May 2011, 01:34
2 In the xy-plane, at what two points does the graph of y = 3 23 Apr 2011, 14:35
7 In the xy-plane at what points does the graph of 4 17 Jan 2010, 12:28
204 In the xy-plane, at what two points does the graph of 17 20 Jan 2007, 23:36
Display posts from previous: Sort by