4
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

$y=2x^2-4x+7$ 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 $x=1$.

$\endgroup$
  • $\begingroup$ Hey I have a prob in the same topic, would you help me out? $\endgroup$ – Kiara Feb 25 '14 at 19:12
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
2
$\begingroup$

$$y=2x^2-4x+7=2\left(x^2-2\cdot x\cdot1+1^2\right)+7-2\cdot1=2(x-1)^2+5$$

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

$\endgroup$

protected by Community Sep 24 '14 at 15:00

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.