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

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Director
Joined: 29 Nov 2012
Posts: 697
If y= (x-1)(x+2), then what is the least possible value of y  [#permalink]

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07 Jul 2013, 07:39
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Question Stats:

53% (01:47) correct 47% (01:50) wrong based on 464 sessions

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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?
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Joined: 02 Sep 2009
Posts: 58401
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
14
15
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 $$x=-\frac{b}{2a}=-\frac{1}{2}$$.

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

Or:

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

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]

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07 Jul 2013, 07:55
6
1
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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##### General Discussion
Director
Joined: 29 Nov 2012
Posts: 697
Re: If y= (x-1)(x+2), then what is the least possible value of y  [#permalink]

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07 Jul 2013, 08:04
2
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]

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07 Jul 2013, 10:41
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]

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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
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Posts: 101
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
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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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29 Oct 2014, 03:40
can someone tell me why it is wrong to just test all the answer values for for x and find the lowest number? That would get you to answer E (-2).

I missed a question once because I assumed that a y= (x equation). Meant x equation was equal to zero. When do we assume it's a quadratic equation?
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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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29 Oct 2014, 13:57
When X has a power 2. Quad comes from the word "square"
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Posts: 77
Re: If y= (x-1)(x+2), then what is the least possible value of y  [#permalink]

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18 Mar 2016, 06:37
1
f(x) = (x-1)(x+2) =x^2+x-2

Taking derivative
F'(x) = 2x+1 =0 =>x=-1/2
f(x) = 1/4 -1/2 -2 = -9/4

Hence B
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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, 03:58
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 $$x=-\frac{b}{2a}=-\frac{1}{2}$$.

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

Or:

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

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
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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, 04:02
1
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.
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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, 07:03
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
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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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05 Jul 2016, 11:52
1
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.
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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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08 Sep 2017, 04:46
Ans is B

y = x^2 + x - 2

dy/dx = 2x+1
for maxima and minima d/dx = 0
x=-1/2

put x = -1/2 in y = x^2 +x -2
y = -9/4

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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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08 Sep 2017, 04:51
Ans is B

Another method apart from derivation

y= x^2 +x -2

lets make it a perfect square
y= x^2 +x +1/4 -1/4 -2
y = {(x+1/2)^2 -1/4} -2

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

therefore we have y =z^2 -9/4
z can be minimum 0 when x=-1/2
so y will be minimum put z=0 so y=-9/4
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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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08 Sep 2018, 21:39
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?

$$y = x^2+x-2$$
By Differentiation - $$x = -1/2$$
Hence $$y = -9/4$$
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Re: If y= (x-1)(x+2), then what is the least possible value of y   [#permalink] 08 Sep 2018, 21:39
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