If y= 4 + (x - 3)^2, then y is lowest when x =

Question

A. 14
B. 13
C. 0
D. 3
E. 4

Chethan92 · Answer

(x-3)^2  is always positive.

minimum value of this expression is 0. So x = 3.

then y = 4.

D is the answer.

KSBGC · Answer

we are looking for the lowest value of y.

y has 2 parts.

1. 4
2.  ( x - 3)^2.

we have nothing to do with 4 but we can curb the value of (x - 3)^2 .

x = 3 will give us the nothing in  (x-3)^2  .

Negative value will change the scenario.  (x-3)^2  always gives us positive value.

The best answer is D.

AnisMURR · Answer

y=4+(x-3)^2

y is the sum of 2 positive numbers

4 is the first one and (x-3)^2 is the second one.

If we are searching the least value of y, we should search the value of x that gives us the least value of (x-3)^2...which is clearly 3.

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Bismarck · Answer

OA:D

y=4+(x-3)^2

{(x-3)^2}\geq{0}

Least value of  (x-3)^2  is  0  at  x=3 .

Least value of  (x-3)^2  will lead to least value of  y  at  x=3 .

Gmat2Cracker · Answer

putting 3 in the eq- y=4+(x-3)^2
y=4 (least)

Ans- (D)

Kinshook · Answer

Given: y= 4 + (x - 3)^2

Asked: y is lowest when x = ?

y= 4 + (x - 3)^2
Since (x-3)^2 >=0
y is lowest when (x-3)^2 =0 or x = 3

IMO D

DemonKing · Answer

I do understand this approach. But I was trying to approach it using calculus (I know it's not required to use calculus on the test) and from what I know, the first derivative of a function is supposed to give the maximum of value of a function. However I tried this and I got x = 3 as my maximizer value of x.

And I looked up elsewhere online that the 2nd derivative equated to zero gives you the minimizer value of x. But I just didn't get any x value as that disappears in the 2nd derivate to leave only 2 as the result.

Would appreciate any clarification on this.

bumpbot · Answer

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If y= 4 + (x - 3)^2, then y is lowest when x =

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If y= 4 + (x - 3)^2, then y is lowest when x =

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