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Find the maximum or minimum value of the quadratic function by completing the squares. Also, state the value of $x$ at which the function is maximum or minimum.

$y=2x^2-4x+7$

$x^2$ has a coefficient of $2$, how should I complete the squares?

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$y=2x^2-4x+7$ therefore $y=2(x^2-2x+7/2)$

completing the square now gives $y=2[(x-1)^2-1+7/2]$

which rearranges to $y=2(x-1)^2+5$

For completing the square, I always divide by a number that ensures I have $1$ as the coefficient of $x^2$.

To get the minimum value there are two options:

1) By looking at the equation after we have completed the square we can see that we will always have the $+5$ term, so we need to minimize the $2(x-1)^2$ term. Because it's square it will never be negative, however, we can make it equal to zero. So, $2(x-1)^2=0$, therefore $(x-1)^2=0$ and $x-1=0$ so $x=1$ is the minimum point.

2) Using differentiation. $y'=4x-4$ then set this equal to zero so $4x-4=0$, $4x=4$ and $x=1$.

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  • $\begingroup$ Hey I have a prob in the same topic, would you help me out? $\endgroup$ – Kiara Feb 25 '14 at 19:12
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    $\begingroup$ @Kiara of course! $\endgroup$ – The Ref Feb 25 '14 at 19:16
  • $\begingroup$ There is another similar question in my book, $y=2x^2-5x-1$, I did it in the same method, but I get this answer: min value is $-3\frac{1}{2}$, occurs when $x$=$\frac{5}{4}$. But the book has an answer like this: min value is $-4\frac{1}{8}$, occurs when $x$=$\frac{5}{4}$. Which one is right? $\endgroup$ – Kiara Feb 25 '14 at 19:21
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    $\begingroup$ you're correct that $x=5/4$, but I'm afraid the book is correct about the y value $2(5/4)^2 - 5(5/4) -1=2(25/16)-25/4 -1=25/8 - 25/4 - 1 = 25/8 - 50/8 - 8/8 = -33/8$ $\endgroup$ – The Ref Feb 25 '14 at 19:29
  • $\begingroup$ Oh got it! Thanks a lot....:D $\endgroup$ – Kiara Feb 25 '14 at 19:30
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$$y=2x^2-4x+7=2\left(x^2-2\cdot x\cdot1+1^2\right)+7-2\cdot1=2(x-1)^2+5$$

or $$2y=4x^2-8x+14=(2x)^2-2\cdot2x\cdot2+2^2+14-4=(2x-2)^2+10$$

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