hi,
It means that Y should be equal to zero.
so a(x-b)^2 = -P
1)a<0
2)P>0
if a(x-b)^2 = -P is going to have a solution then with considering a<0, P has to be positive. and vice versa. so each statement alone is not suffiecient but both together are sufficient.
I will go with C
Co-ordinate geometry
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arashyazdiha
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gmatboost
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Yup. I don't think that graphs of parabolas are tested too much, but it doesn't hurt to know:
P is the y-coordinate of the vertex.
a tells you whether the parabola goes up from the vertex (a > 0) or down (a < 0).
So, if the y-coordinate of the vertex is positive and it heads down, it must cross the x-axis. Either piece of info alone is not enough, since it could start on one side of the y-axis and head away from the y-axis.
P is the y-coordinate of the vertex.
a tells you whether the parabola goes up from the vertex (a > 0) or down (a < 0).
So, if the y-coordinate of the vertex is positive and it heads down, it must cross the x-axis. Either piece of info alone is not enough, since it could start on one side of the y-axis and head away from the y-axis.
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prateek_guy2004
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(1) a < 0 (2) P > 0
the most effective approach here would be to know the value of A..since its given a<0 Its clear that its negative....
For us this is efficient to know the value and range of a.
statement 2 dosent really change anything in terms of intersection.....Insufficient
Answer A
the most effective approach here would be to know the value of A..since its given a<0 Its clear that its negative....
For us this is efficient to know the value and range of a.
statement 2 dosent really change anything in terms of intersection.....Insufficient
Answer A
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gmatboost
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That is incorrect. Knowing that a < 0 is not sufficient. For example:
a = -1, b = 0, P = 1
y = -(x^2) + 1
This equation crosses the y-axis at x = -1 and x = 1
a = -1, b = 0, P = -1
y = -(x^2) - 1
This equation never crosses the y-axis, because the maximum value of [-(x^2) - 1] is -1.
a = -1, b = 0, P = 1
y = -(x^2) + 1
This equation crosses the y-axis at x = -1 and x = 1
a = -1, b = 0, P = -1
y = -(x^2) - 1
This equation never crosses the y-axis, because the maximum value of [-(x^2) - 1] is -1.
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SticklorForDetails
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This is a realistic high-level GMAT question because it does not really test Parabolas. Notice that both of GMATBoost's explanations set b=0 in the examples to make the question into a simple algebra problem. It is fair game on the GMAT to give any function (I have been both quadratic and cubic functions in coordinate geometry questions in GMAC materials) and then ask about points on that curve, because "point on a curve" simply means "solution for x and y in this equation." As arashyazdiha explained, the question really reads,
"In the equation Y= a(x-b)^2 + P, can Y = 0?"
Taken together, we know that it's asking "Can 0 = negative + positive?" to which the answer is a definitive "Yes."
Don't dismiss seemingly "advanced" math topics as un-GMAT-like, but don't spend any time learning such advanced topics really. The GMAT will likely pull from very advanced-looking math content, but all questions are answerable using the limited algebra and strategic approach that we know and love from practice. So on test day, when you see something more "advanced" than you think it should be, you can be confident that there's really a simple algebra or number properties question hidden in there somewhere; all you have to do is find it!
"In the equation Y= a(x-b)^2 + P, can Y = 0?"
Taken together, we know that it's asking "Can 0 = negative + positive?" to which the answer is a definitive "Yes."
Don't dismiss seemingly "advanced" math topics as un-GMAT-like, but don't spend any time learning such advanced topics really. The GMAT will likely pull from very advanced-looking math content, but all questions are answerable using the limited algebra and strategic approach that we know and love from practice. So on test day, when you see something more "advanced" than you think it should be, you can be confident that there's really a simple algebra or number properties question hidden in there somewhere; all you have to do is find it!
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[email protected]
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Y = a(x-b)^2 + P. Can you tell if y has intersects with x axis? (1) a < 0 (2) P > 0
Even if these sort of questions are not tested, I think the answer is specifically E.
Y will intersect the X axis when Y = 0, so hence the question is basically testing that,
whether the equation a (x - b) ^2 + p is equal to 0 or no
from statement 1 we know that the first term is negative and the second term 'p' is positive.
But we do not know the magnitude of these terms. In short the values or both being of the same value.
Eg: if both terms' modulus value is the same then the value of y will be 0.
Hence in short both the statements are not sufficient for the answer to be 0.
Hence E. Please help...
Even if these sort of questions are not tested, I think the answer is specifically E.
Y will intersect the X axis when Y = 0, so hence the question is basically testing that,
whether the equation a (x - b) ^2 + p is equal to 0 or no
from statement 1 we know that the first term is negative and the second term 'p' is positive.
But we do not know the magnitude of these terms. In short the values or both being of the same value.
Eg: if both terms' modulus value is the same then the value of y will be 0.
Hence in short both the statements are not sufficient for the answer to be 0.
Hence E. Please help...
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LalaB
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agree with Gmatboost
here is a graph for those who disagree
https://www.mathwarehouse.com/geometry/p ... rcepts.php
u can change the values of the given parabola to see the the differences
here is a graph for those who disagree
https://www.mathwarehouse.com/geometry/p ... rcepts.php
u can change the values of the given parabola to see the the differences
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