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 October 28th, 2007, 06:51 PM #1 Member   Joined: Sep 2007 Posts: 49 Thanks: 0 extreme values Find the extreme values for this function: y=2x^2-8x+9. I got 2, so that y=1, but the books says '1' is the absolute minimum, but I think it should be local minimum because it didn't tell us the domain. Hope you can help, thanks.
 October 28th, 2007, 09:23 PM #2 Newbie   Joined: Oct 2007 From: Los Angeles Posts: 7 Thanks: 0 the reason why it's an absolute minimum is because this is a parabola with a positive coefficient of the x^2 term. the point (2,1) is the vertex of the parabola.
 October 28th, 2007, 11:09 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,480 Thanks: 2039 If the domain is a real interval (such as all reals) that includes x = 2, the point (2, 1) is a local minimum and the absolute minimum.
October 29th, 2007, 03:21 PM   #4
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Quote:
 Originally Posted by jjinking the reason why it's an absolute minimum is because this is a parabola with a positive coefficient of the x^2 term. the point (2,1) is the vertex of the parabola.
How do you find the vertex?

 October 29th, 2007, 05:15 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,480 Thanks: 2039 y = 2(x − 2)² + 1

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### find the extremebvalue of x^2-8x 9

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