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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, 07:55

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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). _________________

It is beyond a doubt that all our knowledge that begins with experience.

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

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07 Jul 2013, 07:58

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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?

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

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, 22:04

Hello from the GMAT Club BumpBot!

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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, 13:08

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, 02:09

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, 21:38

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).

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