GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 24Get an Extra 30% Off Target Test Prep Plans
08:00 PM EDT
-11:59 PM EDT
Grab 30% off any Target Test Prep GRE plan during our Flash Sale and join Brian and many other students who have used Target Test Prep to score high on the GRE. Just enter the coupon code FLASH30 at checkout to save big.
If y = x2 – 32x + 256, then what is the least possible value
[#permalink]
15 Apr 2018, 03:51
15 Apr 2018, 03:51
Expert Reply
1
Bookmarks
00:00
Question Stats:
57% (00:50) correct
42% (01:23) wrong based on 35 sessions
Hide Show timer Statistics
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
A. 256
B. 32
C. 16
D. 8
E. 0
Drill 2
Question: 12
Page: 512
Question: 12
Page: 512
Re: If y = x2 – 32x + 256, then what is the least possible value
[#permalink]
18 Apr 2018, 02:24
18 Apr 2018, 02:24
1
Among the answer choices \(0\) is the smallest option.
if \(Y = 0\) we have a quadratic eqn which when factorized gives
\((x-16)^2 = 0\)
Therefore when \(x = 16; y= 0\)
option E
if \(Y = 0\) we have a quadratic eqn which when factorized gives
\((x-16)^2 = 0\)
Therefore when \(x = 16; y= 0\)
option E
Re: If y = x2 – 32x + 256, then what is the least possible value
[#permalink]
22 Apr 2018, 12:27
22 Apr 2018, 12:27
Expert Reply
Explanation
First, factor the quadratic equation: \(x2 - 32x + 256 = (x - 16)^2\).
Any quantity squared is either positive or zero. To minimize the expression (x – 16)2 and the value of y, let x = 16, so that y = 0. The answer is choice (E).
First, factor the quadratic equation: \(x2 - 32x + 256 = (x - 16)^2\).
Any quantity squared is either positive or zero. To minimize the expression (x – 16)2 and the value of y, let x = 16, so that y = 0. The answer is choice (E).
