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Expert replies
by ketkoag » Tue Jun 09, 2009 7:09 am
If f(x)=ax^2+3x+1, whether a<0?
1) f(1)=-1
2) The curve represented by the function above doesn't intersect with the line y=2x+3
[spoiler]Please explain how statement 2 is sufficient. OA : D[/spoiler]
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Source: — Data Sufficiency |

by ssmiles08 » Tue Jun 09, 2009 8:23 am
My theory for choice 2 is that when ax^2 + 3x + 1 (irregardless of what sign a is ) is a parabola, the two ends of the parabola are moving to (-infinity , +infinity)

line y=2x+3 also extends towards (-infinity , +infinity), so these two figures are bound to intersect each other inevitably.
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by yogami » Tue Jun 09, 2009 11:26 am
My solution is just intuitive but I am not sure if its right or wrong
This is unlikely to be asked in GMAT cos that first equation is parabolic (cos of it quadratic form). Nevertheless, it seems like f(x) = 0 will yield multiple solutions for x on X axis. However the resulting curve can have symmetric shapes. One curve going upward, the other curve going downward. The one going downward probably has a positive a, and th eone going upward probably has a negative one or vice versa. In any case(2) tells us that the curve does not intersect the line y = 2x + 3. But this line is of the shape / intersecting x axis at its positive side and y axis at its negative side. Which means the curve only has one direction - upward and only one value of a probably supports it. Thus even (2) is sufficient.

OA please and explanation please.
Also if such problems start coming in GMAT then GMAT Is losing its purpose. An arts major in music does not want to deal with parabolas unless he is trying to derive an equation for designing a guitar curve.
200 or 800. It don't matter no more.
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