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

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Re: Coordinate geometry  [#permalink]

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New post Updated on: 23 Apr 2013, 03:31
Zarrolou wrote:
kabilank87 wrote:
In the 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 )


The function is a parabola \(y=x^2+x(a+b)+ab\)

1)Does \(y=x^2-x+ab\) intersec x? It depend on ab.Not sufficient

2) Tells us that point (0,6) is on the parabol, so \(ab=-6\). Does \(y=x^2+x(a+b)-6\) intersect x? It depends on a+b. Not sufficient

1+2) The equation of the parabola is complete \(y=x^2-x-6\), we can tell if it intersect x. Sufficient
C


Hi Zarro ,

In equation no 2 - How you use ab=-6. Is n't ab = +6. Will you please explain the logic behind it ?

Thanks in advance.
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Originally posted by kabilank87 on 23 Apr 2013, 03:27.
Last edited by kabilank87 on 23 Apr 2013, 03:31, edited 1 time in total.
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Re: Coordinate geometry  [#permalink]

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New post 23 Apr 2013, 03:31
1
kabilank87 wrote:
Hi Zarro ,

In equation no 2 - How you replace ab with -6. Will you please explain the logic behind it ?

Thanks in advance.


Sure! Lets take another look at
(2)The graph intersects the y axis at (0, -6 )
We know that the parabola has equation \(y=x^2+x(a+b)+ab\), now if it intersect y in (0,-6) this point is in the parabola, so we just substitute its value (x=0,y=-6) in the equation \(-6=0^2+0(a+b)+ab\) and we find that \(ab=-6\)

Hope it's clear now, let me know
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Re: In the xy plane, at what points does the graph of y=  [#permalink]

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New post 24 Apr 2013, 18:10
This question is a bit tricky , if you know the following it helps :
Any quadratic equation represents a parabola in the x-y domain.
here , the equation is y=(x+a)(x+b) . So the roots of the equation are -a,-b . So it cuts the x axis at -a and -b.

1) a+b = 1 -> Doesn't help since we have one equation with 2 variables.
2) Gives you one set of value for x and y. So you can get an equation in a and b.

Combining 1 and 2 gives the answer to A and B. C wins.
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Re: In the xy plane, at what points does the graph of y=  [#permalink]

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New post 25 May 2014, 07:25
I thought \((x+a)(x+b) = 0\) can be transformed to \(x^2 + ab + 2abx\) since \((x+y)(x+y) = x^2 + y^2 + 2xy\)
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Re: In the xy plane, at what points does the graph of y=  [#permalink]

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New post 16 Jul 2016, 12:08
Top Contributor
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?

IMPORTANT ASIDE ABOUT X-INTERCEPTS:
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

Answer =

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

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New post 17 Jul 2016, 00:14
Answer is C

my thinking is diff here

you are given a quadratic equation of the form x^2+(a+b)x + ab

1)gives u sum of roots.insuff
2) gives u product of roots.iinsuff

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

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New post 14 Oct 2016, 21:12
This boils down to the classic case of sum of roots and product of roots.
the equation is y=x^2+(a+B)x+ab
where a+b is the sum of the roots and ab is the product of the roots.

Statement 1: sum of roots= -1 NS
Statement 2: product of roots=-6 NS

Combining 1 & 2 we get the desired result.
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Re: In the xy plane, at what points does the graph of y=  [#permalink]

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New post 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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Re: In the xy plane, at what points does the graph of y= &nbs [#permalink] 25 Feb 2018, 15:39

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