mejia401 wrote:
Bunuel wrote:
If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =
A. -2
B. -1
C. 0
D. 2
E. None of the above
Kudos for a correct solution.
For a lack of better methods, I lined up each answer choices and input them as a function of \(x\)
A. -2 36<
B. -1 36
C. 0 25D. 2 18E. None of the aboveIMO D.
mejia401 , your solution would have been fine had there not been the last option of "none of the above". You were lucky to get the answer but there might be instances where the actual answer would be "none of the above" and in this case, you will invariably mark the incorrect answer.
\(y = (x – 5)^2 + (x + 1)^2 – 6 = 2x^2-8x+20 = 2x^2-8x+8 + 12\)(try to complete a "\((a \pm b)^2\)")
Thus, \(y = 2(x^2-4x+4) + 12 = 2(x-2)^2 + 12\).
Now, any square of the form \((a-b)^2\) will be MINIMUM, when a=b and the minimum value = 0 ( as for any square , \(x^2 \geq 0\), thus minimum value =0 ).
Based on the discussion above,
minimum value of \(y = 2(x-2)^2 + 12\) , will be when x = 2 and at that value of x, y = 12.