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# Completing the Square in C1 Watch

1. Hey I really need help with this because I can never get it right

Express x2 + 6x + 5 in the form (x+a)2 + b
Write down the co-ordinates of the minimum point for the graph x2 + 6x + 5

I got (x+3)2 + 14 but apparently b should be -4 so why the negative square of 3 rather than the positive?
2. (Original post by Dowel)
Hey I really need help with this because I can never get it right

Express x2 + 6x + 5 in the form (x+a)2 + b
Write down the co-ordinates of the minimum point for the graph x2 + 6x + 5

I got (x+3)2 + 14 but apparently b should be -4 so why the negative square of 3 rather than the positive?
You did this:

(X+3)^2+9 + 5

But if you expand this out do you get the original expression?

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3. x^2+6x+5 goes to (x+3)^2-(3)^2+5, which is (x+3)^2-4. a=3, b=5. Differentiate to get 2x +6, as it is a minimum set the differential to equal 0 thus the gradient at this point on the curve =-3. Then sub -3 into the original equation to get -4 so the minimum co-ordinate is (-3,-4). When completing the square remember it is (x+a/2)^2-(a/2)^2+c, not that we only square the value the negative applies to the value after it is squared (brackets first-BODMAS). Hope that helps all the best for tomorrow. ; )
4. (Original post by kingaaran)
You did this:

(X+3)^2+9 + 5

But if you expand this out do you get the original expression?

Posted from TSR Mobile
so you get x^2 + 6x + 23 with my one and you get the right answer with -9 rather than plus 9, so are you just putting the number that fits the completed square to get the original equation?
5. (Original post by Dowel)
so you get x^2 + 6x + 23 with my one and you get the right answer with -9 rather than plus 9, so are you just putting the number that fits the completed square to get the original equation?
Yeah. Completing the square is all about taking a quadratic and writing it in a different way, without changing it. You see when you have your (x+3)^2, you get x^2+6x and then a +9. This is not what you had in your initial expression, so you must take it away
6. (Original post by kingaaran)
Yeah. Completing the square is all about taking a quadratic and writing it in a different way, without changing it. You see when you have your (x+3)^2, you get x^2+6x and then a +9. This is not what you had in your initial expression, so you must take it away
Thank you so much
7. (Original post by Maz A)
x^2+6x+5 goes to (x+3)^2-(3)^2+5, which is (x+3)^2-4. a=3, b=5. Differentiate to get 2x +6, as it is a minimum set the differential to equal 0 thus the gradient at this point on the curve =-3. Then sub -3 into the original equation to get -4 so the minimum co-ordinate is (-3,-4). When completing the square remember it is (x+a/2)^2-(a/2)^2+c, not that we only square the value the negative applies to the value after it is squared (brackets first-BODMAS). Hope that helps all the best for tomorrow. ; )
Thank you for that as well , is it a coincidence that -3,-4 is so similar to (x+3)^2 -4?
8. (Original post by kingaaran)
Yeah. Completing the square is all about taking a quadratic and writing it in a different way, without changing it. You see when you have your (x+3)^2, you get x^2+6x and then a +9. This is not what you had in your initial expression, so you must take it away
When you complete the square and get the answer in this format e.g. : (x+3)^2 - 4,
how do you know what the coordinates for the minimum and maximum points are?
9. (Original post by SANTR)
When you complete the square and get the answer in this format e.g. : (x+3)^2 - 4,
how do you know what the coordinates for the minimum and maximum points are?
Say you have the function y=(x+3)^2 - 4.

The minimum value of the x-coordinate will be whichever value of x gives us the lowest value of y.

(x+3)^2 can never be negative and so the smallest value it can take is 0. This will happen when x+3=0, which will be when x = -3.

To find the y coordinate, we can then just substitute that in: y=(-3+3)^2 -4 = -4. Hence the minimum point is (-3,-4).
10. (Original post by kingaaran)
Say you have the function y=(x+3)^2 - 4.

The minimum value of the x-coordinate will be whichever value of x gives us the lowest value of y.

(x+3)^2 can never be negative and so the smallest value it can take is 0. This will happen when x+3=0, which will be when x = -3.

To find the y coordinate, we can then just substitute that in: y=(-3+3)^2 -4 = -4. Hence the minimum point is (-3,-4).
Also, how would you work out the maximum point?
11. (Original post by Dowel)
Hey I really need help with this because I can never get it right

Express x2 + 6x + 5 in the form (x+a)2 + b
Write down the co-ordinates of the minimum point for the graph x2 + 6x + 5

I got (x+3)2 + 14 but apparently b should be -4 so why the negative square of 3 rather than the positive?

Dude, the answer is in the form (x+b)2 + c
As a rule of thumb, you subtract b2 to give you the correct c value.

E.g x2 + 6x + 5

Step 1: (x+3)2 -(32) +5

Step 2: (x+3)2 -9 +5 (Notice how I subtract the square of b)

Step 3: (x+3)2 -4 (Simplify c for last step)

KEEP PRACTICING!
12. (Original post by SANTR)
Also, how would you work out the maximum point?
Similar method, you look at how to maximise the function, i.e. y = 4-(x+3)^2. You want (x+3)^2 to be as small as possible, its smallest value is 0 and this happened when x+3=0, x=-3. You then sub this in to find y
13. (Original post by SANTR)
Also, how would you work out the maximum point?
The equation of the graph is a positive parable (x^2). Therefore it doesn't have a maximum only a minimum (curve is at bottom of graph), therefore you can get one value for the co-ordinate of the minimum. You would use the same principle if it was a negative quadratic set the differential to equal 0 then work out the co-ordinate for the turning point. It would be the same with a cubic graph but you would get two values both for a maximum/minimum which by drawing will help to work out which is which.
14. This is quite useful to explain completing the square, especially if the coefficient of x^2 isn't 1

http://www.mathcentre.ac.uk/resource...re2-2009-1.pdf

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