Using the completing-the-square method, find the vertex of the function f(x) = –2x2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point.

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Using the completing-the-square method, find the vertex of the function f(x) = –2x2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point.

Mathematics
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it is a maximum because the leading coefficient|dw:1434844163162:dw| is negative so the parabola opens down
|dw:1434843957618:dw|
Ok:)

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\[ f(x) = –2x^2 + 12x + 5 \]to find the vertex, the first coordinate of the vertex is always \[-\frac{b}{2a}\] which in your case is \[-\frac{12}{2\times (-2)}\]
the second coordinate of the vertex is what you get when you replace \(x\) by the first coordinate
3
Graph opens downward so it would be a maximum point Vertex formula|dw:1434844392512:dw|
yeah that is the first coordinate the second coordinate if \(f(3)\)
Now what do I do?
have a beer you are done
well completing the square \[f(x) = (-2x^2 + 12x) + 5\] factor out -2 \[f(x) = -2(x^2 - 6x) + 5\] now complete the square in x
So is it maximum or minimum??
, find the vertex of the function \[(3,23)\] it is a max have a nice day
Solve x2 + 12x + 6 = 0 using the completing-the-square method.
Thanks friend. I also need help with that one
|dw:1434844578090:dw|
so inside the brackets you need to add (-6/2)^2 so the function is \[f(x) = -2(x^2 - 6x + 9) + 5 -18\] which becomes f(x) = -2(x -3)^2 - 13 now you have completed the square have the equation in vertex form and can answer the question using the required method
|dw:1434844393236:dw|
\[x^2 + 12x + 6 = 0 \\ x^2+12x=-6\] is a start
hahahah you're funny @satellite73 !!!
I see, I see
|dw:1434844713212:dw|
Ok answer the other question please because I'm confused
|dw:1434844498447:dw|
\[x^2+12x=-6\]takes two steps to complete the square what is half of 12?
6
right, and what is \(6^2\_?
oops what is \(6^2\)?
36
so go right from \[x^2+12x=-6\] to \[(x+6)^2=-6+36\] or \[(x+6)^2=30\]
then take the square root of both sides, don't forget the \(\pm\) and get \[x+6=\pm\sqrt{30}\]
subtract 6 and you are done
so is it 6 plus or negative radical 30?
no
SUBTRACT 6 to solve for \(x\)
|dw:1434844993983:dw|
nope
|dw:1434845032667:dw|
|dw:1434845026665:dw|
yeah that one

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