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kpkanupriyakhmi
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5LB Q18 page 927

by kpkanupriyakhmi Sat Mar 28, 2015 12:20 pm

Diagram
then :-Which quadrant, if any, contains no point (x, y) that satisfies the inequality y (x - 3)2 - 1?
(A) I
(B) II
(C) III
(D) IV
(E) All quadrants contain at least one point that satisfies the given inequality.
My doubt:-
Why this:-x - 3 < 1 or x - 3 > -1
AND
I didn't understand the graphical solution how can one interpret that the curve will not pass through third Quadrant?
n00bpron00bpron00b
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Re: 5LB Q18 page 927

by n00bpron00bpron00b Sun Mar 29, 2015 12:34 pm

Is plugging in/elimination a more efficient approach here, Tommy ?


Which quadrant, if any, contains no point (x, y) that satisfies the inequality y >= (x - 3)2 - 1?


a) 1st quadrant (+,+)
(3,2) - 2>= -1 - True - choice a out

b) 2nd quadrant (-,+)
(-3, 45) - 45>=(-6)^2 - 1
45 >= 36-1
45>=35 - True - choice b out

d) 4th quadrant (+,-)
(3,-1) -> -1 >=(3-3)^2 - 1
-1 >= -1 - True - choice d is out

Between choices (c) 3rd quadrant (-,-) and (d) All quadrants contain at least one point that satisfies the given inequality.

Choice (c) will always fail
LHS will always be negative and RHS positive

so choice (c) - 3rd quadrant
tommywallach
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Re: 5LB Q18 page 927

by tommywallach Mon Mar 30, 2015 6:58 pm

Hey Guys,

First off, kp, please use ^2 for squaring.

And yes, as Noob said, you DO NOT want to try and graph this. Instead, just think about the equation:

y ≥ something squared - 1

Well, (x-3)^2 will always be positive (or zero). If x is positive, then we could get (x-3) very close to 0 (if x = 2.9, for example), in which case the right side of the equation would be negative.

But if x is negative, the right side of the equation HAS to be positive, so you can never have a situation where both x and y are negative.

You can get here through "logic" or plugging in numbers. Just don't graph.

-t
kpkanupriyakhmi
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Re: 5LB Q18 page 927

by kpkanupriyakhmi Mon Mar 30, 2015 7:24 pm

And why this in solution
Why this:-x - 3 < 1 or x - 3 > -1?
tommywallach
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Re: 5LB Q18 page 927

by tommywallach Fri Apr 03, 2015 5:15 pm

That explanation is positing a hypothetical situation about y (that it's negative), then solving for x. You don't have to do that, it's just explaining things as deeply as possible.

Just do the question the ways we suggested--plugging in numbers or conceptualizing. : )

-t