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by vinod_ece66 » Thu Jun 12, 2008 9:05 am
Keep substituting values for x and note the corresponding values of y. You will see that there is a line which will pass through all the three quadrants expect 2nd quadrant.

ex: x=0 y=-5
x=1 y=0
x=2 y=3
x=3 y=4
x=-1 y=4
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by AleksandrM » Thu Jun 12, 2008 9:32 am
I did the following:


y = -(9 - 6x + x^2) + 4

y = -9 + 6x - x^2 + 4

x^2 + y = 6x - 5

I just ignored the x^2 and worked with y = mx + b

y-intercept is -5 [plot that]

the slope is 6/1 [plot that, up and down (e.g. up 6 and 1 over to the right and down 6 and 1 over to the left - when you draw a line, you'll see that the only quadrant that the line does not pass through is QII)]
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by durgesh79 » Thu Jun 12, 2008 9:40 am
plug in values and eliminate options.

First E is out, the graph has to pass through atleast one of quardrant

Check Q1, with x = 3,
y = 4, Options A and D are out.

with x = -1, y is -ve -----> Q3, option C is out

B is answer.
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by babachal » Thu Jun 12, 2008 2:39 pm
hi dude, for this question you have to learn how to draw parabola from quadratic equation, answer is asking you curve not line!

the answer is true, it doesnt intersecet quadrant II, but how?
y=-(3-x)^2+4=-x^2+6x-5
when y=o, then x1=1, x2=5 where the curve intersect x-axis:
when x=0, then y=-5 this where one side of curve intersects y-axis.
now we have to find peak point point of parabola which is equal to -b/2a from ax^2+bx+c=0, here b=6 and a =-1 then peak point is 3 from y point, the maximum or minium point of curve.

x1=1, x=5
y=-5
peak point is y=3
now try to draw the parabola,

it begins from y=3
left side intersects (1,0) and continues till intersects again (0,-5)
right side intersects (5,0) then continues unlimitedly,
if imagine this graph, it passed 1, III, and IV quadrants only, not II quadrant!

please visit the following website,
https://www.purplemath.com/modules/grphquad2.htm
try to understand if you cant write me private message, i will respond you as soon as I can.

if u have any questions, let me know!
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by Ian Stewart » Thu Jun 12, 2008 5:01 pm
I think there are a few good approaches to this problem, some of which were suggested above. If you plot a few points (by plugging in a few values for x), you can work out quite quickly what the curve looks like. If you see that this equation represents a parabola, and you know how to draw parabolas, even better (though that isn't required knowledge on the GMAT). It's also possible to look at this purely algebraically. It's easier to rewrite the right side:

y = 4 - (3-x)^2

If x is positive, (3-x)^2 could be as small as zero, and could be as large as you like. So if x is positive, y might be positive (but no larger than 4), or might be negative. So the curve can be in quadrants I and IV. If x is negative, however, notice that (3-x)^2 must be larger than 9; the smaller ("more negative") you make x, the larger this gets. When you subtract something larger than 9 from 4, you must get a negative number. So if x is negative, y must also be negative, and the curve can never be in quadrant II.
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by spanlength » Fri Jun 13, 2008 3:09 am
Hi friends , as i see the posts related to the above question , I try to analyze the problem in a diiferent way:

A knowledge of parabola is helpful but i would use the transformation of the axis .Shiftng the origin to (3,4) we have a basic parabola eq. and the above equation will change to x^2 = ay, where a is a constant.
We know that the original curve is not passing through origin so when we shift back it will make intercept in +ve x axis and -ve yaxis , there by not passing through II quadrant.

Hope this useful to future readers , to analyze the problems in a different way.

Any comments !!
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