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

Question

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

Bunuel · Accepted Answer

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. https://gmatclub.com/forum/download/file.php?id=10339 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.

Zarrolou · Answer

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).

fozzzy · Answer

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

mau5 · Answer

y = x^2+x-2 	o x^2+x-(2+y)=0 	o  For real values of x, the Discriminant (D)\geq{0} 	o 1^2-4*1* \geq{0}	o

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

B.

sunaimshadmani · Answer

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

usre123 · Answer

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

angelfire213 · Answer

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?

usre123 · Answer

When X has a power 2. Quad comes from the word "square"

Vinayprajapati · Answer

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

rhine29388 · Answer

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

Bunuel · Answer

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.

rhine29388 · Answer

Always a pleasure learning from you kudos to your post

LakerFan24 · Answer

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.

sahilvijay · Answer

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

ANSWER IS B

sahilvijay · Answer

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

rahul16singh28 · Answer

y = x^2+x-2 
By Differentiation -  x = -1/2 
Hence  y = -9/4

Kinshook · Answer

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

y = x^2 + x - 2 = (x+1/2)^2 -2 - 1/4 = (x+1/2)^2 - 9/4 
Least possible value of (x+1/2)^2 = 0

Least possible value of y = - 9/4

IMO B

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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

Prep Toolkit

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