Hello,
Can you please tell me how to solve this:
How many times does graph y = ax^2 + bx + c intersect the x-axis?
1) a > 0
2) c < 0
OA: C
I was also just wondering if these type of questions can be tested on the GMAT?
Thanks a lot,
Sri
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How many times does y = ax^2 + bx + c intersect the x-axis?
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gmattesttaker2
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Hi Sri,
The concepts in this DS question are quite rare, but possible, on the GMAT. I wouldn't put too much time or energy into these sorts of graphing questions though, as you're not likely to see this on the GMAT.
That having been said, there are some Number Properties behind this type of graphing:
1) A graph of a quadratic will be a "U" shape (either "open up" or "open down").
If the "a" variable is positive, then the graph opens "up"; if the "a" is negative, then the graph opens "down"; if the "a" is 0, then you have a line.
2) The graph of a quadratic will intersect the X axis at 0, 1 or 2 points.
3) The y-intercept of this line (the "c" variable) is important because it anchors the graph at a point and then the graph either opens "up" or "down"
Fact 1: "a" is positive
This tells us that the graph opens "up", but we don't know where the y-intercept is:
If "c" is positive, then the answer is 0
If "c" is 0, then the answer is 1
If "c" is negative, then the answer is 2
Fact 1 is INSUFFICIENT
Fact 2: "c" is negative
This tells us the y-intercept is negative, but we don't know if the graph opens "up" or "down":
If "a" is positive, then the answer is 2
If "a" is 0, then the answer is 1
If "a" is negative, then the answer is 0
Fact 2 is SUFFICIENT
Combined, we know the "a" is positive (so the graph opens "up") and the "c" is negative (so the y-intercept is negative). This graph would have 2 intercepts.
Combined, SUFFICIENT
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
The concepts in this DS question are quite rare, but possible, on the GMAT. I wouldn't put too much time or energy into these sorts of graphing questions though, as you're not likely to see this on the GMAT.
That having been said, there are some Number Properties behind this type of graphing:
1) A graph of a quadratic will be a "U" shape (either "open up" or "open down").
If the "a" variable is positive, then the graph opens "up"; if the "a" is negative, then the graph opens "down"; if the "a" is 0, then you have a line.
2) The graph of a quadratic will intersect the X axis at 0, 1 or 2 points.
3) The y-intercept of this line (the "c" variable) is important because it anchors the graph at a point and then the graph either opens "up" or "down"
Fact 1: "a" is positive
This tells us that the graph opens "up", but we don't know where the y-intercept is:
If "c" is positive, then the answer is 0
If "c" is 0, then the answer is 1
If "c" is negative, then the answer is 2
Fact 1 is INSUFFICIENT
Fact 2: "c" is negative
This tells us the y-intercept is negative, but we don't know if the graph opens "up" or "down":
If "a" is positive, then the answer is 2
If "a" is 0, then the answer is 1
If "a" is negative, then the answer is 0
Fact 2 is SUFFICIENT
Combined, we know the "a" is positive (so the graph opens "up") and the "c" is negative (so the y-intercept is negative). This graph would have 2 intercepts.
Combined, SUFFICIENT
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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Brent@GMATPrepNow
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I agree with Rich; I'd say that this question is out of scope, but only barely.
The math syllabus in the OG13 mentions parabolas (the same formed by the equation y = ax² + bx + c) only ONCE (on page 139). On page 139, we see how one can find points on a parabola by plugging in various x-values to find the corresponding y-values. This is a pretty unsophisticated way to graph an equation, and it does not require us to know how the values of a, b and c affect the graph of y = ax² + bx + c (which we all learned in high school).
Another way to determine how many times the graph of an equation intersects the x-axis is to set that equation to zero. So, we get ax² + bx + c = 0. We can use counter-examples to show why statements 1 and 2 are not sufficient, but showing/proving that the combined statements are sufficient would, in my opinion, require skills that are just beyond what the GMAT requires.
Cheers,
Brent
The math syllabus in the OG13 mentions parabolas (the same formed by the equation y = ax² + bx + c) only ONCE (on page 139). On page 139, we see how one can find points on a parabola by plugging in various x-values to find the corresponding y-values. This is a pretty unsophisticated way to graph an equation, and it does not require us to know how the values of a, b and c affect the graph of y = ax² + bx + c (which we all learned in high school).
Another way to determine how many times the graph of an equation intersects the x-axis is to set that equation to zero. So, we get ax² + bx + c = 0. We can use counter-examples to show why statements 1 and 2 are not sufficient, but showing/proving that the combined statements are sufficient would, in my opinion, require skills that are just beyond what the GMAT requires.
Cheers,
Brent
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For any equation of the form y = ax² + bx + c, the DISCRIMINANT is equal to b² - 4ac.gmattesttaker2 wrote:
How many times does graph y = ax^2 + bx + c intersect the x-axis?
1) a > 0
2) c < 0
OA: C
If the discriminant is POSITIVE, then the graph of the equation intersects the x-axis TWICE.
If the discriminant is ZERO, then the graph of the equation intersects the x-axis ONCE.
If the discriminant is NEGATIVE, then the graph of the equation DOES NOT INTERSECT the x-axis.
Question stem, rephrased: Is b² - 4ac positive, zero, or negative?
Statement 1: a > 0
Test one case that satisfies BOTH statements.
Case 1: If b=1, a=1, and c=-1, then b² - 4ac = 1² - (4*1)(-1) = 5, which is positive.
Test one case that satisfies ONLY STATEMENT 1.
Case 2: If b=1, a=1, and c=1, then b² - 4ac = 1² - 4*1*1 = -3, which is negative.
INSUFFICIENT.
Statement 2: c < 0
Case 1 also satisfies statement 2.
In Case 1, the discriminant is positive.
Test one case that satisfies ONLY STATEMENT 2.
Case 3: If b=1, a=-1, and c=-1, then b² - 4ac = 1² - (4)(-1)(-1) = -3, which is negative.
INSUFFICIENT.
Statements combined:
Since the square of a value cannot be negative, b² ≥ 0.
Since a>0 and c<0, 4ac < 0.
Thus:
b² - 4ac = nonnegative - negative = nonnegative + positive = positive.
SUFFICIENT.
The correct answer is C.
I wouldn't worry too much about this problem.
Questions about parabolas appear on the GMAT only rarely.
That said, my solution above combines two concepts discussed briefly in the OG13:
In its discussion of quadratics on page 125, the OG13 mentions that the number of SOLUTIONS depends on the sign of b² - 4ac.
In its discussion of parabolas on page 139, the OG13 mentions that the x-interecepts occur when the quadratic is EQUAL TO 0.
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Hello Rich, Brent, Mitch,
Thank you all for your solutions and for sharing your thoughts on this question.
Best Regards,
Sri
Thank you all for your solutions and for sharing your thoughts on this question.
Best Regards,
Sri
