In the xy-plane shown, the shaded region consists of all points that

Dear dgboy765,
I'm happy to help.

Here are the basic ideas. Consider the graphs of the form y = [some expression of x]. These can include
a) oblique lines, y = mx + b (e.g. y = (3/7)x + (5/7))
b) parabolae: e.g. y = x^2, or y = x^2 - 4
c) higher powers of x: e.g. y = x^5 - x^3
d) square roots: y = sqrt(x - 3)
e) all kinds of other exotic curves: y = 2^x, or y = (1 + x)/(1 - x), or etc.
For the purpose of this discussion, the nature of that variety doesn't matter. What I am going to say applies to all graphs of the form y = [expression of x]. I will use the parabola y = x^2 - 4 as my example graph, but what I am saying about this graph applies to any graph of the form y = [expression of x]

The first may be obvious: all the values (x, y) that satisfy the equation y = x^2 - 4 must lie exactly on the graph of y = x^2 - 4. More generally, the graph of any equation is the set of all ordered pairs that satisfy this equation.

Now, think about any point on that line. Imagine we take an (x, y) that live on the graph, and we change the y-coordinate to make it bigger. That would result in a new point that is not on the graph but above the graph. This corresponds to the inequality y > x^2 - 4. Any ordered pair that satisfies this inequality is somewhere above the graph.

Similarly, any ordered pair that satisfies the inequality y < x^2 - 4 lies somewhere below the graph.

In this question, notice that for point (a, b), a is the x-coordinate and b is the y-coordinate. The shaded region shown in the picture is below the x-axis, so y < 0, and it is above the parabola y = x^2 - 4, so it is entirely defined in terms of these two inequalities:
y < 0 and y > x^2 - 4
The prompt already specified that b < 0, and that takes care of the first inequality. Notice that statement #2 is just the second inequality written in terms of a & b instead of x & y. That's why it's sufficient. Once we specify both inequalities, we have specified the shaded region entirely.

On a side note, notice that y = x, the line through the origin with a slope of 1, is the line that contains all points with equal x- and y-coordinates. The inequality y > x are all the points above y = x, and the inequality x > y are all the points below y = x.

Does all this make sense?
Mike
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They explain the concept of how the xy co-ordinate plane is divided into two parts by a line. The parabola does the same. It divides the plane into two regions:
y > x^2 - 4x
y < x^2 - 4x

Stmnt 2 gives us that (a, b) lies in the region y > x^2 - 4x which is the region inside the parabola (if you want to verify, check (2, 1). It satisfies y > x^2 - 4x.
Since the question stem tells us that b < 0, so we are looking at a point below the x axis. The region inside the parabola below x axis is the shaded region. So the point must lie on it.
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Hi,
In very simple terms to solve this Q..

The type of parabola etc is amplified by mike in the above post...
A parabola of Quadratic equation will have a minimum or a maximum value depending on the coeff of \(x^2\)..
here it is positive, so the parabola will be open upwards and will have a mininmum value at\(x=\frac{-b}{2a}\) or 4/2=2.. and the value is \(2^2-4*2=-4\)...

lets see the statements now..

(1) \(0 < a < 4\)..
nothing about b....
so point (a,b) can be anywhere depending on value of b
at a=3.99, b can be -4, so will be outside the graph or at some point inside..
and at a=2, b can be -3.99, it will be inside the graph..
so insuff..

(2) \(a^2 - 4a < b\)
the moment you see this equation, its similarity with the original equation y=x^2 - 4x should strike you..
we substitute a and b as x and y in the eq we get b=a^2-4a...
since the equation \(b=a^2-4a\) is that of the the line..
a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it...
so suff..
you can test this with, say at the x axis..
at a=4, b=0..
\(a^2-4a=b... 4^2-4*4=b=0\)..
so if \(a^2-4a<0, 0<a<4\) satisfies the condition for within the shaded portion and so suff..
the moment \(a^2-4a>0\), a>5 or a<0 on x axis, and this point will be outside the parabola..
hope it helped you in some way
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Hi Mike
Actually, I can not get the concept behind your explanation.
Please explain more taking into account the basic concepts since I did not master the Coordinate concepts.
Thank you in advance.

Dear hatemnag,
I'm happy to respond.

My friend, before I answer your question, I am going to challenge you. I am going to challenge you to ask a better, more thorough question. See this blog article:
Asking Excellent Questions

You see, the question you asked was very vague and general. Vague questions are poor questions. It's perfectly fine that you don't understand some basic concepts about Coordinate Geometry and that you want to learn more--in fact, it's wonderful that you are asking for help! The trouble is, I have absolutely no idea what you already know and what you need. Your vague question completely leaves me in the dark.

An excellent question would involve making explicitly clear, exactly and specifically, what parts you understand and what parts you don't understand. For example, you could go through each point I make in my Jan. 27, 2016 post above, and tell me exactly what you understand and exactly what confuses you about each item.

Of course, a more specific, more detailed question, will help me respond to you more effectively, but what students often fail to understand: I recommend crafting an excellent question, not for me, but for you! The process of putting all the effort into writing an excellent question will be tremendously helpful to you: by explaining all this, you will force yourself to make connections and have realizations, and all this effort will prime your mind so that my more detailed response will be that much more helpful for you. It's much harder to produce an excellent question, and, in fact, all the effort it takes to craft an excellent question is actually an investment in your own understanding and learning. When a student poses an excellent question to the teacher, it's a total win-win scenario. This is precisely why asking excellent questions is one of the habits of excellence.

Therefore, my friend, I am going to challenge you to write an excellent question. Craft the highest quality question you can, making explicitly and specifically clear precisely which parts you understand and don't understand about coordinate geometry. It's fine to have a ton of questions, that's great, but it's important for both of us to appreciate the basic ideas that you do understand, because all learning is based on what you already grasp.

Does all this make sense?

Mike
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hi there, mikemcgarry and chetan2u can someone help me out? I am taking the GMAT in 8 days (my practice CATs have been in the 700 range Q49, V40), but this question came as a bit of a shock to me. Note, I have not read either of your explanations yet because this concept looks new to me. What is "a" and "b" referring to? Is it the a and b within the quadratic equation (ax^2+bx+c=0)? For the most part, the only questions I have seen related to graphs in my practice is lines (y=mx+b). I will read your explanations shortly, but first, is there a gap in my understanding of the basics? I think I remember from grade school this U shape being a "Parabola." Is that correct? If so, what about Parabola's do we need to know for the GMAT?

Hi gzimmer,

a and b refer to a point in the plane where a is on x-axis and B on y-axis..
You are correct that such equation refers to parabola.
Also parabola if asked would be basic, as for instance one can see the similarities in two equations- one in Q and other in statement II.
And that would also mean you are likely to be doing well in Quant till then.

Now with just 8 days left, just do the basic and do not spend too much time on topics rarely seen ..
All the best.
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Hi,
I solved it this way:

remember that stem says b<0.

st1. clearly insufficient because say y= -5 then it is not in shade region but for y greater that -4 it is in region.

st2. you can rewrite it as a^2-4a+4<b+4 (add 4 to both sides) then LHS is (a-2)^2 and RHS is b+4 ---> (a-2)^2<a+4 as LHS is something always positive then it can not be smaller than something negative thus b+4 must be greater than 0, therefore we have -4<b<0 hence 0<b+4<4. ---> (a-2)^2<4 ---> |x-2|<2 --> -2<x-2<2 --> 0<x<4. it is in the region. suff.
B

In Statement 1 couldn't Point B be anything? For instance -100000 ? little bit confused why you picked -3.99 and and -4 for b ? or am I missing something ?


Im sure Im missing something or the questions is just very basic

I mean from statement 1: no information about b at all --> so clearly insufficient.

Statement 2: it explicitly tells us that a,b lies above the graph and it is given that b<0

Hi VeritasPrepKarishma mikemcgarry shashankism Bunuel

Can you suggest flaw in below approach:

Given: Question has asked me whether a point (a,b) lies on shaded curve.
I hope there is no discrepancy between whether the shaded curve includes the parabola.
As per mine understanding it does not.

Essentially the parabola is in quadrant IV and hence (x,-y) should be co-ordinates of points in this quadrant.

St 1: 0<a<4 -> I simply took in random values between 0 to 3.9 and found that since x is positive st 1 is sufficient.
After reading the solutions and above explanation found that for two different values of a/x, say 1 or 8 I am not alway
getting positive values of y hence st 1 is insufficient.

St 2: at first glance the equation translates to x^2 - 4x < y since (a,b) have to in shaded region but for these how did we
intuitively assume that (a,b) is on curve? this query is especially pertaining to above explanation by mikemcgarry

As per my understanding it seems we have put an upper gap on parabola by saying b<0 in question stem. Any view on this?
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hi..

since b is not given and no relation exists between a and b, b can take any value..
you are correct in your understanding...two points given were just to illustrate that b could be inside or outside graph
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I actually spent a lot of time on this simple question.

Here is the mistake I did for anyone who is still confused: I assumed that "a" is on x2-2x. Therefore, when I saw condition (1), 0<a<4, I started plugging in values for a in X2-2x . This would have worked if the assumption was stated; however, at no point did the question say that a is on function x2-2x. Therefore, we can not determine "b" from "a" because we are not given any info that ties the two point together.

Hi, thanks for your explanation. I have a doubt

In condition B - I understand that a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it.

However, the question asks us about the shaded region. Whereas a^2-4a< b will include all area inside the parabola - both above x axis and below x axis. this to ensure that the point lies under the x axis we will need Condition A (Which we can't prove/ derive from Condition B)

Therefore I feel Answer should be C. Please let me know if I am missing out on something

Hello

The point you make is valid, that a^2 - 4a < b will include all area inside the parabola, both above x axis and below x axis. But its also mentioned in the question stem that b < 0. Since y coordinate is negative, the area that we have to consider is the one below x-axis only.

Hi Guys,

I have spent too many minutes on this questions which is also on OG quant review 2019 DS01613, I am still not able to get through the questions ask and the explanation given here. I think i have a thick brain it seems, Can Bunuel or VeritasKarishma help me in understanding the question stem and theory/concept behind it. Thanks

If you are getting confused with the normal way of solving but you are great at visualizing and drawing coordinate geometry, here's the video solution for you

https://gmatquantum.com/official-guides ... nt-review/

Statement 1- tells us nothing much. Point can lie inside or outside the parabola in range 0<x<4.
Insufficient

Statement 2- Locus of point (a,b) is \(y>x^2-4x\) , where y<0. In other words locus of the (a, b) lies below the y-axis and inside the parabola \(y=x^2-4x\).

Sufficient.

VeritasKarishma - could you please help in this question. Difficult to understand

Posted from my mobile device

I have talked about it here: https://gmatclub.com/forum/in-the-xy-pl ... l#p2191933
Let me know if you still have doubts.
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