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21 Jun 2017, 13:59
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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
a. -3
b. -9/4
c. -2
d. -3/2
e. 0
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21 Jun 2017, 14:25
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If y= (x-1)(x+2) = x^2 + x - 2
Now when
x=0, y = 0 + 0 - 2 = -2
x=-1/2, y=1/4 - 1/2 - 2 = -9/4 = -2.25
x=-1/3, y=1/9 - 1/3 - 2 = -8/3 = -2.67
x=-1/4, y=1/16 -1/4 - 2 = -35/16 = -2.18
x=-1, y = 1 - 1 -2 = -2
Of the available options, Option B is smallest possible value given in the answer options
The least value of y occurs when x=-1/2
Now when
x=0, y = 0 + 0 - 2 = -2
x=-1/2, y=1/4 - 1/2 - 2 = -9/4 = -2.25
x=-1/3, y=1/9 - 1/3 - 2 = -8/3 = -2.67
x=-1/4, y=1/16 -1/4 - 2 = -35/16 = -2.18
x=-1, y = 1 - 1 -2 = -2
Of the available options, Option B is smallest possible value given in the answer options
The least value of y occurs when x=-1/2
21 Jun 2017, 15:19
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we get the equation for y as y = x^2 +x-2
This is in the form ax^2 +Bx +c, the least value of the expression is when x=-b/2a
here a=1, b=1 and c=-2
Least value for y is when x=-b/2a => = -1/2
Substituting this value
y=(-1/2)^2+(-1/2)-2
=1/4-1/2-2
=-1/4-2
=-9/4
Option B
This is in the form ax^2 +Bx +c, the least value of the expression is when x=-b/2a
here a=1, b=1 and c=-2
Least value for y is when x=-b/2a => = -1/2
Substituting this value
y=(-1/2)^2+(-1/2)-2
=1/4-1/2-2
=-1/4-2
=-9/4
Option B
