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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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Updated on: 05 Jun 2019, 07:31

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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)

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 \((x+a)(x+b)=0\).

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.

Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)
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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 \((x+a)(x+b)=0\).

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 [ 3.61 KiB | Viewed 132537 times ]

Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)
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03 May 2011, 18:33

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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.

Answer (C).
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)
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Updated on: 26 Oct 2011, 07:17

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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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I agree. However once we get x=a,b I used this info in st 1

a+b = -1

Since a=x and b=x

X+x = -1 x= -1/2

I know what I’m doing is not right but theoretically I cannot seem to understand why this approach is incorrect

Thanks

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In addition to what Bunuel said above, think in this way too:

When we have (x + a)(x + b) = 0, x is a variable and a and b are some constant values.

We know that x must take one of two values: -a or -b x must be either -a or -b so that the product becomes 0.

When you say a = -x, and b = -x, you are assuming that a and b are equal and their value is some constant -x (which is not the case)

e.g. (x + 2)(x - 1) = 0 implies x is either -2 or 1. Can I say that -2 = x and 1 = x so (-2 + 1) = x + x = -1? This gives us x = -1/2 but we know that x is either -2 or 1. Then this is not possible. Our assumption here is that a and b are equal which they needn't be.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)
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25 May 2011, 03:26

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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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I agree. However once we get x=a,b I used this info in st 1

a+b = -1

Since a=x and b=x

X+x = -1 x= -1/2

I know what I’m doing is not right but theoretically I cannot seem to understand why this approach is incorrect

Thanks

Posted from my mobile device

x-intercepts of a parabola y = (x + a)(x + b) are x = -a and x = -b. This does not mean that -a = -b. For example, The x-intercepts of y = (x + 1)(x + 2) are x = -1 and x = -2.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)
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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

Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)
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24 Dec 2009, 10:37

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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.

Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)
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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 \((x+a)(x+b)=0\).

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.

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

Answer C.
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Re: In the xy-plane, at what two points does the graph of y=(x+a)(x+b)
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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.

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

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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)
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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.

Answer = C

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.

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

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