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# In the xy-plane, at what two points does the graph of y=(x+a)(x+b)

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

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Updated on: 05 Jun 2019, 07:29
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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)

Originally posted by misterJJ2u on 21 Jan 2007, 00:36.
Last edited by Bunuel on 05 Jun 2019, 07:29, edited 3 times in total.
Renamed the topic and edited the question.
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In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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Updated on: 05 Jun 2019, 07:31
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1
93
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)

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.

Hope it's clear.
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Originally posted by Bunuel on 22 Oct 2009, 13:28.
Last edited by Bunuel on 05 Jun 2019, 07:31, edited 2 times in total.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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

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22 Oct 2009, 13:45
1
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

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

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22 Oct 2009, 13:51
2
(2) The graph intersects y-axis at (0,-6) --> x=0, y=-6 --> y=x^2+x(a+b)+ab --> -6=0^2+0(a+b)+ab --> ab=-6.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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24 Dec 2009, 10:37
2
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.
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In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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15 Oct 2010, 01:05
1
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

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

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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.

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

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25 Dec 2010, 08:49
8
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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.

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 [ 3.61 KiB | Viewed 132085 times ]
Hope it's clear.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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03 May 2011, 10:21
1
1
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.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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03 May 2011, 18:33
5
5
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.

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

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25 May 2011, 03:26
1
1
To find x-intercepts, you set y=0 and solve. The question asks for the x-intercepts of y = (x+a)(x+b). Setting y=0, the question is asking when (x+a)(x+b) = 0. Since one of the factors on the left side must be 0, we find we have intercepts when x = -a and x = -b.

Statement 1 tells us the sum of a and b, which doesn't let us find their values alone.

Statement 2 gives us a point on the curve; if we plug that point into our equation, what we get must be true. So -6 = (0+a)(0+b), and ab = -6. So Statement 2 gives us the product of a and b, which isn't enough to find their individual values.

Together, we know that a+b = -1 and ab = -6. That is, we know the sum and product of two numbers. Notice this is what happens every time you factor a quadratic: you look for two numbers with a certain product and a certain sum. For example, if you were to factor the quadratic x^2 - x - 6, you'd look for two numbers which multiply to -6 and add to -1. So if a+b = -1 and ab = -6, then our solutions for a and b must be -3 and 2, in some order. Since our intercepts are at -a and -b, the intercepts are at x=3 and x=-2, so the information is sufficient.

Notice we can't figure out which value is equal to a and which is equal to b, but we don't need to do that to answer the question; we just need to know where the two points are, and not which point is a and which point is b.
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In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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24 Jul 2011, 21:40
blog 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).

The question stem asks us the points where the graph intersects the X axis. What does 'intersect the X axis' imply? It implies that Y = 0 at the points where the graph intersects the X axis because on the entire X axis, Y = 0.

Y = (X+a)(X+b) will intersect the X axis when Y = 0.
(X + a)(X + b) = 0
We will get two values for the X co-ordinate (since there are two points). The two values will be -a and -b. So basically, we are looking for the values of a and b. (If you remember your high school Math, this is a quadratic equation so its graph is an inverted U. It intersects the X axis at two points - Just to give you some perspective.)

Statement 1: a + b = -1
a and b can take infinite different values. Not sufficient alone.

Statement 2: The graph intersects the Y-axis at (0,-6)
'Intersect at Y axis' implies that X = 0.
When X = 0, we are given that Y co-ordinate is -6.
Let's put this in the given equation: Y = (X+a)(X+b)
-6 = ab
a and b can take infinite different values. Not sufficient alone.

Together, we know that a + b = -1 and ab = -6
Two different equations and two unknowns. We will be able to find the values of a and b. Hence together they are sufficient. (We don't actually need to find the values but you can see that a and b will be -3 and 2 giving us -a and -b as 3 and -2.)

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

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Updated on: 26 Oct 2011, 07:17
2
1
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.
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Originally posted by blink005 on 26 Oct 2011, 03:04.
Last edited by blink005 on 26 Oct 2011, 07:17, edited 2 times in total.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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26 Feb 2012, 23:50
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.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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27 Feb 2012, 00:03
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.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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30 Apr 2012, 12:17
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.

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

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16 Jul 2016, 12:08
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JimmyWorld 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)

Target question: At which two points of the graph does y=(x+a)(x+b) intersect the x-axis?

Let's examine the point where a line (or curve) crosses the x-axis. At the point of intersection, the point is on the x-axis, which means that the y-coordinate of that point is 0. So, for example, to find where the line y=2x+3 crosses the x-axis, we let y=0 and solve for x. We get: 0 = 2x+3
When we solve this for x, we get x= -3/2.
So, the line y=2x+3 crosses the x-axis at (-3/2, 0)

Likewise, to determine the point where y = (x + a)(x + b) crosses the x axis, let y=0 and solve for x.
We get: 0 = (x + a)(x + b), which means x=-a or x=-b
This means that y = (x + a)(x + b) crosses the x axis at (-a, 0) and (-b, 0)
So, to solve this question, we need the values of a and b

Aside: y = (x + a)(x + b) is actually a parabola. This explains why it crosses the x axis at TWO points.

Now let's rephrase the target question...
REPHRASED target question: What are the values of a and b?

Statement 1: a + b = -1
There's no way we can use this to determine the values of a and b.
Since we can answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The line intercepts the y axis at (0,-6)
This tells us that when x = 0, y = -6
When we plug x = 0 and y = -6 into the equation y = (x + a)(x + b), we get -6 = (0 + a)(0 + b), which tells us that ab=-6
In other words, statement 2 is a fancy way to tell us that ab = -6
Since there's no way we can use this information to determine the values of a and b, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
Statement 1 tells us that a+b = -1
Statement 2 tells us that ab = -6

Rewrite equation 1 as a = -1 - b
Then take equation 2 and replace a with (-1 - b) to get: (-1 - b)(b) = -6
Expand: -b - b^2 = -6
Set equal to zero: b^2 + b - 6 = 0
Factor: (b+3)(b-2) = 0
So, b= -3 or b= 2

When b = -3, a = 2 and when b = 2, a = -3
In both cases, the two points of intersection are (3, 0) and (-2, 0)
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

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

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25 Feb 2018, 15:39
Hi All,

Like most questions on the GMAT, this question can be approached in a number of different ways. There's actually a great Algebra pattern/shortcut built into this question:

We given the equation Y = (X+A)(X+B) and we're asked at what 2 points the graph will intersect with the X-axis. This essentially comes down to the A and B. If we know their values, then we can answer the question. It's also worth noting that since we're multiplying, you can "flip-flop" the values of A and B and you'd have the same solution.

For example: (X+1)(X+2) is the same as (X+2)(X+1)…...

1) A + B = -1

There's no way to determine the exact values for A and B with this information.

IF...
A = 0, B = -1
A = 100, B = -101
Etc.
Different numbers for A and B would lead to different solutions.
Fact 1 is INSUFFICIENT

2) The graph intersects the Y-axis at (0, -6).

Now we have one of the points on the graph. Plugging it into the original equation gives us….

-6 = (0+A)(0+B)
-6 = AB

We have the same situation as in Fact 1: more than 1 possible solution.
A = 1, B = -6
A = 2, B = -3
Etc.
Fact 2 is INSUFFICIENT

Combined, we have…
A + B = -1
AB = -6

Here's where things get interesting. This is a "system" of equations, so we CAN solve it...the "catch" is that the answers would "flip flop":

If you did do the math, you'd have
A = -3, B = 2
OR
A = 2, B = -3

Since this is a graphing question, these two options provide the SAME solution.
Combined, SUFFICIENT

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# Rich Cohen

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Joined: 12 Jul 2018
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Location: India
Schools: ISB '20, NUS '21
GMAT 1: 420 Q26 V13
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)  [#permalink]

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13 Dec 2018, 09:18
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)   [#permalink] 13 Dec 2018, 09:18

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