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

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23 Apr 2013, 02:27
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 ?

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Last edited by kabilank87 on 23 Apr 2013, 02:31, edited 1 time in total.

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23 Apr 2013, 02:31
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kabilank87 wrote:
Hi Zarro ,

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

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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24 Apr 2013, 17:10
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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25 May 2014, 06:25
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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16 Jul 2016, 11:08
Expert's post
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?

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

[Reveal] Spoiler:
C

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

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16 Jul 2016, 23:14

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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14 Oct 2016, 20:12
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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16 Oct 2017, 07:49
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Re: In the xy plane, at what points does the graph of y=   [#permalink] 16 Oct 2017, 07:49

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