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

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If y= (x-1)(x+2), then what is the least possible value of y [#permalink] New post 07 Jul 2013, 06:39
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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?
[Reveal] Spoiler: OA

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Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink] New post 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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Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink] New post 07 Jul 2013, 06:58
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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.
Image

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}.

Therefore y_{min}=(-\frac{1}{2}-1)(-\frac{1}{2}+2)=-\frac{9}{4}


Answer: B.

Or:

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.
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Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink] New post 07 Jul 2013, 07:04
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Here's how I did this one...

x^2 + x - 2

we can complete the square since (x+a)^2 = x^2 + 2ax + a^2

Here 2ax = x >>a=\frac{1}{2}

(x+1/2)^2 - 1/4 - 2 = y subtracting 1/4 since 1/4 will be added when its squared.

(x+1/2)^2 - 9/4 = y

y = -9/4
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Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink] New post 07 Jul 2013, 09:41
Expert's post
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^2+x-2 \to x^2+x-(2+y)=0 \to For real values of x, the Discriminant(D)\geq{0} \to 1^2-4*1*[-(2+y)]\geq{0}\to

y+2\geq{-\frac{1}{4} }\to y\geq{-\frac{9}{4}}

B.
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Re: If y= (x-1)(x+2), then what is the least possible value of y   [#permalink] 07 Jul 2013, 09:41
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