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Bigcnee
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#21
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#21
(Original post by bono)
lol. i did the question that way!!!! got it right..

but my point was, it first asks me to complete the square for the equation. i just dont see how i can find the least val. from compelting the square.

and yes, i agree, i did it like this, the minimum turning point for this quadratic equation, but the question is on about complete. square, then "hence"...
You probably didn't get any marks for it then.
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Nima
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#22
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#22
(Original post by Expression)
Minimum value occurs when gradient is equal to zero, ie the turning point on a simple quadratic.

f(x) = x^2 - 12x + 40
f'(x) = 2x - 12

f'(x) = 0 => 2x - 12 = 0 => 2x = 12 => x = 6

So minimum value of f(x) occurs when x = 6.

Sub x=6 into f(x)

6^2 - 12(6) + 40
= 36 - 72 + 40
= 4

So minimum value of f(x) is 4.


[edit]Looks at Bhaal85's answer, and can't believe that I overlooked using "Complete the Square" to find the co-ordinates of the vertex[/edit]
i get ur method, because we find the min. value by finding the x coordinate of the vertex, then substictue into quadratic......
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Nima
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#23
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#23
(Original post by Bigcnee)
You probably didn't get any marks for it then.
i didnt do an exam, i just saw it now on my desk lol.
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Expression
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#24
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#24
(Original post by bono)
lol. i did the question that way!!!! got it right..

but my point was, it first asks me to complete the square for the equation. i just dont see how i can find the least val. from compelting the square.

and yes, i agree, i did it like this, the minimum turning point for this quadratic equation, but the question is on about complete. square, then "hence"...
The Completed Square form of a Quadratic ax^2 + bx + c , is

a(x+p)^2 + q and it is fact, that you should accept, that this gives the points of the vertex (the turn in the graph), which would sit at (-p, q).

From this form, you can deduce that (assuming the vertex gives a minimum), the minimum is q, and occurs when x = -p.
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Bigcnee
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#25
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#25
(Original post by bono)
i didnt do an exam, i just saw it now on my desk lol.
Oh. Consider yourself lucky then!
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Expression
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#26
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#26
(Original post by Bigcnee)
You probably didn't get any marks for it then.
I agree with Bigcnee to a certain extent here, if it asked you to use a specific method to find the minimum then you will have got 1 mark for finding the minimum as an end result.

If however, the question somewhat spoon fed you:

a) Find the completed square form of f(x)

b) Hence find the minimum value.

Then you may have gotten a few more marks.
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Bhaal85
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#27
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#27
(Original post by bono)
i get ur method, because we find the min. value by finding the x coordinate of the vertex, then substictue into quadratic......
Complete the square:

(x-6)^2+4

therefore least value = 4 at x=6

To find the vertex (lowest point) you set the thing in the brackets to = 0

Therefore: X-6=0, making X=6. That gives you the X Co-ord. Now you need to obtain the Y Co-ord. Which is 4, since there is no outside coeffecient, its just (6,4).
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Nima
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#28
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#28
(Original post by Expression)
The Completed Square form of a Quadratic ax^2 + bx + c , is

(x+p)^2 + q and it is fact, that you should accept, that this gives the points of the vertex (the turn in the graph), which would sit at (-p, q).

From this form, you can deduce that (assuming the vertex gives a minimum), the minimum is q, and occurs when x = -p.
thankyou ever so much 4 jogging my memory mate, i remmeber now.
(3 months ago lol.)

thanks.
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Expression
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#29
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#29
(Original post by Bigcnee)
You probably didn't get any marks for it then.
I agree with Bigcnee to a certain extent here, if it asked you to use a specific method to find the minimum then you will have got 1 mark for finding the minimum as an end result.

If however, the question somewhat spoon fed you:

a) Find the completed square form of f(x)

b) Hence find the minimum value.

Then you may have gotten a few more marks, although where is says hence, it suggests that your answer for a) is a BIG clue in the answer for b) !
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Nima
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#30
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#30
(Original post by Expression)
I agree with Bigcnee to a certain extent here, if it asked you to use a specific method to find the minimum then you will have got 1 mark for finding the minimum as an end result.

If however, the question somewhat spoon fed you:

a) Find the completed square form of f(x)

b) Hence find the minimum value.

Then you may have gotten a few more marks, although where is says hence, it suggests that your answer for a) is a BIG clue in the answer for b) !
one thing.

lookin at the mark scheme for this, there is only one mark for finding the least value.

therefore, if u put "the lest value for the given quadratic equation is 4", then u get the mark.

therefore, if in doubt, u could always do it by finding the min turnin point, using differentiation (i.e. on scrap paper) then put ur answer down and get all the marks.

or, u cud just do a bit f revision and remember how to find it after completing the square, as expression reminded me.
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