If y=4+(x-3)^2, then y is least when x
A)14
B)13
C)0
D)3
E)4
Answer is D.
I don't understand what the question is asking. Can someone explain, please?
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Brent@GMATPrepNow
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We want to minimize the value of 4+(x-3)^2shanice wrote:If y=4+(x-3)^2, then y is least when x
A)14
B)13
C)0
D)3
E)4
Answer is D.
I don't understand what the question is asking. Can someone explain, please?
To do this, we need to minimize the value of (x-3)^2
Notice that (any number)^2 > 0
So, the smallest value possible of (any number)^2 is 0
(x-3)^2 = 0 when x=3.
So, the value of 4+(x-3)^2 is minimized when x=3.
Answer = D
Cheers,
Brent
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To MINIMIZE y, we must MINIMIZE the value of 4 + (x-3)².shanice wrote:If y=4+(x-3)^2, then y is least when x
A)14
B)13
C)0
D)3
E)4
Answer is D.
I don't understand what the question is asking. Can someone explain, please?
Since the square of a value cannot be negative, the least possible value of (x-3)² is 0.
(x-3)² = 0 when x=3.
Thus, y will be its least possible value when x=3:
y = 4 + (x-3)² = 4 + (3-3)² = 4.
If x is any value other than 3, the value of y will INCREASE.
The correct answer is D.
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truplayer256
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y = 4 + (x - 3)^2
Note that y is always going to increase or stay the same no matter what since (x - 3)^2 is ALWAYS going to be positive. Therefore, the only obvious choice for the least value of y is 4 and this occurs when x = 3.
If you have taken calculus before, you can also use differentiation to come up with your answer:
dy/dx = 2(x - 3) = 0 => x = 3
This means that a minimum occurs when x = 3.
The best choice is D.
Note that y is always going to increase or stay the same no matter what since (x - 3)^2 is ALWAYS going to be positive. Therefore, the only obvious choice for the least value of y is 4 and this occurs when x = 3.
If you have taken calculus before, you can also use differentiation to come up with your answer:
dy/dx = 2(x - 3) = 0 => x = 3
This means that a minimum occurs when x = 3.
The best choice is D.
