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Doubt on quadratics.

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Doubt on quadratics.

by bblast » Sun Sep 04, 2011 7:06 am
From Gmat prep :

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)Graph intersects y axis at (0, -6)

I could easily solve this; answer -> C, combining both we get (x-3)(x+2) = 0; hence intersection points are 3, -2.


From knewton :
If f(x) = ax² + bx + c, and a does not equal 0, at what point does the graph of the function f(x) intersect the y-axis?

1>The graph of f(x) intersects the x-axis exactly twice, at (-6, 0) and (-2, 0)
2>a = 2
I chose A since :
we just use statement 1 to write an equation - > f(x)= (x+6)(x+2). Now to find where this intersects the Y axis we set x = 0. thus answer f(0) = 12. hence sufficient.

But, correct answer is C again. https://www.knewton.com/blog/gmat/challe ... intersect/

need someone to point the difference between the 2 questions ?
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by knight247 » Sun Sep 04, 2011 8:01 am
In case of the knewton problem
We don't know the value of a, whether -ve or +ve. Because the sign of the square term indicates whether the parabola is a peak or a valley(sorry if this sounds too technical)

(1)if the graph intersects x axis at (-6, 0) and (-2, 0), it could happen with +ve or -ve value of a. In each case the y intersept will be different. U can't substitute the value of a and b as -6 and -2 in the same equation. (-6, 0) and (-2, 0) are the co ordinates of the parabola's x intercepts...They are not a and b. Its done in a different way if y=f(x) = ax² + bx + c then u need to put x=-6 and y=0 and get one quadratic equation and then put x=-2 and y=0 and get a second quadratic eqn. When you do this, you will get two quadractic eqns with 3 variables and we can't solve two quadratic equations with 3 variables each. We need to isolate the value of C so not sufficient.

(2)if a=2 then we get an indicator that the graph is a peak but we still can't get the y intercept.Insufficient

Combining both we get 2 quadratic equations with two variables i.e. b and c you can solve for both. Then you'll have the values of both a,b and c. Put these values in the equation y=f(x) = ax² + bx + c we get y=4x²+5x+7......(just speculating, these are not the real values of a b and c). And since we have to find the y co-ordinate i.e. the minimum value of the parabola(the only point at which the parabola intersects the y axis) we put x=0 and we get the value of y. Hence C
Last edited by knight247 on Sun Sep 04, 2011 10:19 am, edited 1 time in total.
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by bblast » Sun Sep 04, 2011 8:14 am
Thanks Knight,

I get the point on peak and valley. Long back I read the thory in MGMAT guides, there is also some logic that the larger the magnitude of A, the narrower the curve.

One doubt still remains. After solving the first question from Gprep, i assumed that the reverse is also true,

(x-3)(x+2). if i set x =0 in this equation, i get {f(0) = point of intersection at y = -6}. this is the info given in statement 2. Thats the theory I was following to solve the knewton problem and hence made the error.

Could u shed some light on this ?


Off subject : Looks from the pic like u are having a drink somewhere in the shacks of goa ?
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by knight247 » Sun Sep 04, 2011 8:16 am
bblast,
also watch this video

https://www.youtube.com/watch?v=RQHqd050FqE

it should clarify any doubts you have about solving parabolic equations
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by bblast » Sun Sep 04, 2011 9:24 am
Thanks KNight, to add on, i got the crisp difference between the two questions from a math genius in below lines :


"In the Knewton question the equation of graph will be y = a(x+6)(x+2) because coefficient of x^2 is 'a'(not 1). So, f(0) = 12a. So, we need value of 'a'. Hence, it is C.
But in the GMAT Prep question, a=1, so that is the difference between these two questions."
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by knight247 » Sun Sep 04, 2011 10:05 am
Hey bblast,

Hope you watched the youtube video. My doubts also got cleared after watching it. If you have an equation like 3x+4y=7 then this is a straight line. Meaning, at some point or the other it is going to touch the y axis and at some point or the other it is going to touch the x axis. So putting x=0 will give you the y intercept and putting y=0 will give you the x intercept. And the two points you get are the ONLY two times that the line will ever intercept either of the two axes. That was for a straight line. This funda can't really be applied to parabola's as they are a little more complex. Let me solve the gmat prep problem for you.

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)Graph intersects y axis at (0, -6)

y=(x+a)(x+b)
y=x²+x(a+b)+ab(have skipped a few steps in the middle)

(1)a+b=-1 so putting this value in the above eqn we have
y=x²-x+ab

If we put x=0 we get y=ab and y=0 then x=-ab. If a+b=-1 the values of a and b could be anything. So we don't know at which two points it intersects the x axis. So INsufficient

2)Graph intersects y axis at (0, -6). Now we already know that the parabola is a valley along y axis. So it only intersects the y axis at its minimum point. Not enough to find both its x intercepts. So insufficient.

Combining both we get y=x²-x+ab with the minimum value as (0, -6)
Putting x=0 and y=-6 in the above quadratic eqn we have
-6=ab
ab=-6
so eqn can be re written as y=x²-x-6 now on both the x intercepts y will be equal to zero. So putting y=0 we get x²-x-6=0. And we will get the two values of x as 3 and -2...
So, the two points of intersection are (3,0) and (-2,0). Phew..I'm exhausted...lol
Hope this clarifies ur doubt. Lemme know if i've made any mistakes and if u need more clarification.

And, yes that pic is in Goa. Beautiful beach not very far from Anjuna at a beach side restaurant called Boom Shankar. I've promised myself a 4-5 month vacation to Goa once I kill it on the GMAT. Fingers crossed.lol. Do you travel around Goa a lot? I'm guessing ur from Mumbai

Made a slight change to the post
Last edited by knight247 on Mon Sep 05, 2011 8:18 am, edited 2 times in total.
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by knight247 » Sun Sep 04, 2011 10:20 am
Also made a change to my post on the knewton problem. Check it out.
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by GMATGuruNY » Sun Sep 04, 2011 11:19 am
bblast wrote:From Gmat prep :

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)Graph intersects y axis at (0, -6)


The graph intersects the x axis when y=0.
Substituting y=0 into y=(x+a)(x+b):
0 = (x+a)(x+b).
x= -a or x= -b.
Question rephrased: What are a and b?

Statement 1: a+b = -1.
No way to solve for a and b.
Insufficient.

Statement 2: The graph intersects the y axis at (0, -6).
Substituting x=0 and y=-6 into y=(x+a)(x+b):
-6 = (0+a)(0+b)
-6 = ab
No way to solve for a and b.
Insufficient.

Statements 1 and 2 combined:
Substituting a= -b-1 into ab= -6:
(-b-1)b = -6.
-b²-b+6 = 0.
b²+b-6 = 0.
(b-2)(b+3) = 0.
b=2 or b=-3.
If b=2, then a= -b-1= -2-1 = -3.
If b=-3, then a= -b-1 = -(-3)-1 = 2.
Both cases yield the same combination of values: 2 and -3.
Whether a=2 and b=-3 or a=-3 and b=2 is immaterial; the graph will cross the x axis at 2 and -3.
Sufficient.

The correct answer is C.
From knewton :
If f(x) = ax² + bx + c, and a does not equal 0, at what point does the graph of the function f(x) intersect the y-axis?

1. The graph of f(x) intersects the x-axis exactly twice, at (-6, 0) and (-2, 0).
2. a = 2
The graph intersects the y axis when x=0.
Substituting x=0 into y=ax²+bx+c:
y = a(0²) + b(0) + c
y = c.
Question rephrased: What is c?

Statement 1: The graph of f(x) intersects the x-axis exactly twice, at (-6, 0) and (-2, 0).
Substituting x=-6 and y=0 into y = ax²+bx+c:
0 = a(-6)² + b(-6) + c
0 = 36a-6b+c.

Substituting x=-2 and y=0 into y = ax²+bx+c:
0 = a(-2)² + b(-2) + c
0 = 4a-2b+c.

3 variables, but only 2 distinct linear equations.
No way to determine the value of c.
Insufficient.

Statement 2: a=2.
No way to determine the value of c.
Insufficient.

Statements 1 and 2 combined:
Substituting a=2 into 0=36a-6b+c and 0=4a-2b+c:
0 = 72-6b+c and 0 = 8-2b+c.
2 variables, 2 distinct linear equations, so we can solve for c.
Sufficient.

The correct answer is C.
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by bblast » Sun Sep 04, 2011 10:41 pm
Thanks Mitch.

@knight - you made a small mistake in the last step.

y=(x+a)(x+b)
Graph intersects y axis at (0, -6).

thus :
point (0,6) is a point at which x = 0. The function at this point will be f(0)
f(x) = (x+a)(x+b)
f(0) = (0+a)(0+b)
also f(x)= y ; thus y = ab
thus ab = -6. (which is the info provided by b). Final equation is x²-x-6 = (x-3)(x+2)


Off topic : I already have a Goa trip planned next month after submitting ISB app post GMAT. :)
Cheers !!

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