In the xy plane, at what two points does the graph of y = (x a x b) intersect the x axis

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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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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 20 Jan 2007, 23:36.
Last edited by Bunuel on 05 Jun 2019, 06:29, 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]

Updated on: 05 Jun 2019, 06:31

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.Answer: C.Hope it's clear. _________________

Originally posted by Bunuel on 22 Oct 2009, 12:28.
Last edited by Bunuel on 05 Jun 2019, 06:31, 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]

21 Jan 2007, 01:49

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]

22 Oct 2009, 12:45

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

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]

22 Oct 2009, 12:51

(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]

24 Dec 2009, 09:37

y=(x+a)(x+b) when y=0To 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=0So let's expand, and we have x^2 + (a + b)x + ab = 0So 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]

15 Oct 2010, 00:05

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

Answer is (c)

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

25 Dec 2010, 03:49

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.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.

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

25 Dec 2010, 07:49

gautamsubrahmanyam wrote:

Bunuel wrote:

Hope it's clear.

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 207738 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]

03 May 2011, 09:21

y=(x+a)(x+b) = x^2+(a+b)x+abthis is parabola equation. intersects x-axis means y=0st 1--> a+b = -1x^2-x+ab=0 cannot solve the equation; not sufficientst 2--> graph intersects y-axis at -6i.e. ab = -6x^2+(a+b)-6=0 cannot solve the eq, not sufficient.Both together, x^2-x-6=0Solving 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]

03 May 2011, 17:33

udaymathapati wrote:

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 -bHence 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 = -1Two 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) [#permalink]

25 May 2011, 02:26

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]

24 Jul 2011, 20:40

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) = 0We 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 = -1a 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 = aba and b can take infinite different values. Not sufficient alone.Together, we know that a + b = -1 and ab = -6Two 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) [#permalink]

Updated on: 26 Oct 2011, 06:17

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=-12. 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=-1Insufficient2.-6=(0+a)(0+b)ab=-6Insufficient.1+2a+b=-1ab=-6a=-3, b=2a=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.

Originally posted by blink005 on 26 Oct 2011, 02:04.
Last edited by blink005 on 26 Oct 2011, 06:17, edited 2 times in total.

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

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.

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

26 Feb 2012, 23:03

mourinhogmat1 wrote:

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]

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 = -1ab = -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. Let me know if you have any further questions.

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

Updated on: 03 Jul 2020, 07:08

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 = -6Since 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 = -1Statement 2 tells us that ab = -6Rewrite equation 1 as a = -1 - b Then take equation 2 and replace a with (-1 - b) to get: (-1 - b)(b) = -6Expand: -b - b^2 = -6Set equal to zero: b^2 + b - 6 = 0Factor: (b+3)(b-2) = 0So, b= -3 or b= 2When 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 SUFFICIENTAnswer: C _________________

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

25 Feb 2018, 14:39

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

Final Answer:

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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, 08:18

If you prefer the explanation in video format: