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Re: y = x^2 - 16x + 64 [#permalink]
GreenlightTestPrep wrote:
Carcass wrote:
\(y = x^2 - 16x + 64\)

Quantity A
Quantity B
The least value of y
0


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Given: \(y = x^2 - 16x + 64\)
Factor right side to get: \(y = (x-8)^2\)

If we SQUARE any number squared, the result will always be greater than or equal to zero.
In this case, we can minimize y by letting x = 8

If \(x = 8\), then \(y=(x-8)^2=(8-8)^2=(0)^2=0\)

We get:
QUANTITY A: 0
QUANTITY B: 0

Answer: C

Cheers,
Brent


If we SQUARE any number squared, the result will always be greater than or equal to zero.
Is don't understand this line. Explanation pls!
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Re: y = x^2 - 16x + 64 [#permalink]
1
For a no \(x\), positive or negative or zero

\(x^2 ≥ 0\)

You can try as well. Take any value and square it. :thumbsup:

SivhHarish wrote:
If we SQUARE any number squared, the result will always be greater than or equal to zero.
Is don't understand this line. Explanation pls!
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Re: y = x^2 - 16x + 64 [#permalink]
1
SivhHarish wrote:
If we SQUARE any number squared, the result will always be greater than or equal to zero.
Is don't understand this line. Explanation pls!


Sorry, I shouldn't have added that second squared.
It should read: If we SQUARE any number, the result will always be greater than or equal to zero
Examples:
3² = 9, and 9 ≥ 0
(1.5)² = 2.25, and 2.25 ≥ 0
0² = 0, and 0 ≥ 0
(-2)² = 4, and 4 ≥ 0
(-0.1)² = 0.01, and 0.01 ≥ 0
etc
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Re: y = x^2 - 16x + 64 [#permalink]
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