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If y = x^2 − 32x + 256, then what is the least possible value of y ?

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Bunuel
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If y = x^2 − 32x + 256, then what is the least possible value of y ? [#permalink] New post   14 Mar 2019, 00:18
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If y = x^2 − 32x + 256, then what is the least possible value of y ?

A. 256
B. 32
C. 16
D. 8
E. 0
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E
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pushpitkc
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If y = x^2 − 32x + 256, then what is the least possible value of y ? [#permalink] New post   14 Mar 2019, 00:23
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\(y = x^2 − 32x + 256 = (x - 16)^2\)

At x=16, the value of y is 0

Therefore, the least value of y from the options available is 0(Option E)
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Re: If y = x^2 − 32x + 256, then what is the least possible value of y ? [#permalink] New post   14 Mar 2019, 03:57
Bunuel wrote:
If y = x^2 − 32x + 256, then what is the least possible value of y ?

A. 256
B. 32
C. 16
D. 8
E. 0

y = x^2 − 32x + 256

(x-16)*(x-16)=0
x=16
so at x=16
we get y=0
IMO E
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Re: If y = x^2 − 32x + 256, then what is the least possible value of y ? [#permalink] New post   18 Mar 2019, 18:49
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Bunuel wrote:
If y = x^2 − 32x + 256, then what is the least possible value of y ?

A. 256
B. 32
C. 16
D. 8
E. 0


Since x^2 − 32x + 256 = (x - 16)(x - 16) = (x - 16)^2, we see that the least possible value of y is 0 (occuring at x = 16).

Answer: E
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Re: If y = x^2 − 32x + 256, then what is the least possible value of y ? [#permalink] New post   09 Jul 2020, 10:51
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Bunuel wrote:
If y = x^2 − 32x + 256, then what is the least possible value of y ?

A. 256
B. 32
C. 16
D. 8
E. 0


Asked: If y = x^2 − 32x + 256, then what is the least possible value of y ?

y = x^2 − 32x + 256 = (x-16)^2
Least possible value of y = 0 when x = 16

IMO E
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Re: If y = x^2 − 32x + 256, then what is the least possible value of y ? [#permalink] New post   26 Feb 2021, 22:06
\(ax^2+bx+c\)
When a is positive the parabola has the vertex at the bottom (and when negative the vertex is at the top)
Here \(a=1\) and positive. The minimum of the parabola will be at the vertex.

The formula for the x value of the vertex (whether positive or negative) is \(-b/2a\). We can then use this value to solve for the y value of the vertex.

\(-b/2a=-(-32)/2(1)=32/2=16\)
Substitute 16 for x
\(y=16^2-32(16)+256\)
Take out the common factor
\(y=16(16-32+16)=16(0)=0\)
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Re: If y = x^2 − 32x + 256, then what is the least possible value of y ? [#permalink]
26 Feb 2021, 22:06
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