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Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

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07 Jul 2013, 06:55

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fozzzy wrote:

If y= (x-1)(x+2), then what is the least possible value of y?

A. -3 B. -9/4 C. -2 D. -3/2 E. 0

Any alternative solutions?

The function is a parabola with a positive "a" coefficient

The roots are x=1 and x=-2, for values between 1 and -2 it will have negative values (E is out)

\(y=x^2+x-2\), you can try to insert values (starting from the least) to see if it could be the answer. Example \(-3=x^2+x-2\) or \(x^2+x+1=0\) =>Impossible \(-\frac{9}{4}= x^2+x-2\) or \(x^2+x+\frac{1}{4}=0\) => \(x=-0.5\) valid => CORRECT

Or approach #2:

Given the roots -2 and 1, the x of the vertex will be the middle point => \(x=-0.5\)

And the least value will be the y coordinate of the vertex (plug \(x=-0.5\) into the equation).
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If y= (x-1)(x+2), then what is the least possible value of y?

A. -3 B. -9/4 C. -2 D. -3/2 E. 0

Any alternative solutions?

\(y= (x-1)(x+2)=x^2+x-2\).

Theory: Quadratic expression \(ax^2+bx+c\) reaches its extreme values when \(x=-\frac{b}{2a}\). When \(a>0\) extreme value is minimum value of \(ax^2+bx+c\) (maximum value is not limited). When \(a<0\) extreme value is maximum value of \(ax^2+bx+c\) (minimum value is not limited).

You can look at this geometrically: \(y=ax^2+bx+c\) when graphed on XY plane gives parabola. When \(a>0\), the parabola opens upward and minimum value of \(ax^2+bx+c\) is y-coordinate of vertex, when \(a<0\), the parabola opens downward and maximum value of \(ax^2+bx+c\) is y-coordinate of vertex.

Examples: Expression \(5x^2-10x+20\) reaches its minimum when \(x=-\frac{b}{2a}=-\frac{-10}{2*5}=1\), so minimum value is \(5x^2-10x+20=5*1^2-10*1+20=15\).

Expression \(-5x^2-10x+20\) reaches its maximum when \(x=-\frac{b}{2a}=-\frac{-10}{2*(-5)}=-1\), so maximum value is \(-5x^2-10x+20=-5*(-1)^2-10*(-1)+20=25\).

Back to the original question: \(y= (x-1)(x+2)=x^2+x-2\) --> y reaches its minimum (as \(a=1>0\)) when \(x=-\frac{b}{2a}=-\frac{1}{2}\).

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

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27 Oct 2014, 21:04

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Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

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28 Oct 2014, 12:08

Can we also use b^2 - 4ac to determine the valid minimum value, as thats what I did.. I got 0 as the answer for B and I thought 0 means invalid so went for C. I want some clarification here, first if b^2-4ac=0, it means eq has no real soln?? Or my approach for this question was wrong ! Thanks in advance

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

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29 Oct 2014, 01:09

I did it this way The minimum (or max, if the coef is -a) should lie half way between the two roots which are -2 and 1. halfway between that is -3/2. Now put in -3/2 in place of x: (-3/2 -1)(-3/2 +2) = -9/4

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

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17 Mar 2016, 20:38

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

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04 Jul 2016, 02:58

Bunuel wrote:

fozzzy wrote:

If y= (x-1)(x+2), then what is the least possible value of y?

A. -3 B. -9/4 C. -2 D. -3/2 E. 0

Any alternative solutions?

\(y= (x-1)(x+2)=x^2+x-2\).

Theory: Quadratic expression \(ax^2+bx+c\) reaches its extreme values when \(x=-\frac{b}{2a}\). When \(a>0\) extreme value is minimum value of \(ax^2+bx+c\) (maximum value is not limited). When \(a<0\) extreme value is maximum value of \(ax^2+bx+c\) (minimum value is not limited).

You can look at this geometrically: \(y=ax^2+bx+c\) when graphed on XY plane gives parabola. When \(a>0\), the parabola opens upward and minimum value of \(ax^2+bx+c\) is y-coordinate of vertex, when \(a<0\), the parabola opens downward and maximum value of \(ax^2+bx+c\) is y-coordinate of vertex.

Examples: Expression \(5x^2-10x+20\) reaches its minimum when \(x=-\frac{b}{2a}=-\frac{-10}{2*5}=1\), so minimum value is \(5x^2-10x+20=5*1^2-10*1+20=15\).

Expression \(-5x^2-10x+20\) reaches its maximum when \(x=-\frac{b}{2a}=-\frac{-10}{2*(-5)}=-1\), so maximum value is \(-5x^2-10x+20=-5*(-1)^2-10*(-1)+20=25\).

Back to the original question: \(y= (x-1)(x+2)=x^2+x-2\) --> y reaches its minimum (as \(a=1>0\)) when \(x=-\frac{b}{2a}=-\frac{1}{2}\).

Use derivative: \(y'=2x+1\) --> equate to 0: \(2x+1=0\) --> \(x=-\frac{1}{2}\) --> \(y_{min}=(-\frac{1}{2}-1)(-\frac{1}{2}+2)=-\frac{9}{4}\)

Answer: B.

Hope it's clear.

Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them

Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them

If you know how to use any of the advanced techniques of course you can use it. Personally I've never seen a GMAT question requiring something like this. For example, I've never seen a geometry question requiring trigonometry, every GMAT geometry question can be solved without it.
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Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

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04 Jul 2016, 06:03

Bunuel wrote:

rhine29388 wrote:

Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them

If you know how to use any of the advanced techniques of course you can use it. Personally I've never seen a GMAT question requiring something like this. For example, I've never seen a geometry question requiring trigonometry, every GMAT geometry question can be solved without it.

Always a pleasure learning from you kudos to your post

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

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05 Jul 2016, 10:52

i did NOT do this problem like the other posters did on here...way too complicated. i just plugged in #s.

start w/ x=1. if x=1, (1-1)=0, so A/C= 0. x=2: (2-1)(4)>0, so 1 is greatest positive # we can do. ok. x=-1: (-2)(1) = -2. Now this is lowest #. Elim 0. x=-2: (-3)(0) = 0. Not quite. x=-3: (-4)(-1) = positive. Not going in right direction. **Answer must be -2<x<0. x= -1/2: (-1/2-2/2)(-1/2+4/2)= (-3/2)(3/2)= -9/4. Good! x= -3/2: (-3/2-2/2)(-3/2+4/2)= (-5/2)(1/2) = -5/4. * -9/4 = -2.25; -5/4= -1.25. -9/4 = correct. A/C B = correct.

gmatclubot

Re: If y= (x-1)(x+2), then what is the least possible value of y
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05 Jul 2016, 10:52

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