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# GCSE maths - vertex of graphs watch

1. ok, my teacher was going on about how to do this by putting the equation in complete the square form and then using that 2 find out the vertex of the graph but i've kinda 4gotten how 2 do this and need some help. i'm not sure if this will come up or not but it'll probz be the last question. so how do u find out the vertex of a quadratic graph is basically wat i need an answer to. thanks
2. (Original post by urban_flavaz)
ok, my teacher was going on about how to do this by putting the equation in complete the square form and then using that 2 find out the vertex of the graph but i've kinda 4gotten how 2 do this and need some help. i'm not sure if this will come up or not but it'll probz be the last question. so how do u find out the vertex of a quadratic graph is basically wat i need an answer to. thanks
by vertex wat do u mean?
using complete the square to give the poistive and neg points of the graph?
Its jsut lyk factorisin..what dont u get completing the square?
3. If you could give us an example of some sorts it would be better.
Vertex could mean the points where the line intercepts the y axis??
4. (Original post by jamal1425)
If you could give us an example of some sorts it would be better.
Vertex could mean the points where the line intercepts the y axis??
say for eg u have a question saying:

"what are the minimum/max points on the graph 2x^2 + 8x - 5
you would do (x+4)(x+4) which gives you x^2 + 8x + 16 but we want -5 so we would take off 21 thus giving:
(x+4)^2 -21
this then makes
x= +/- the square root of -21 -4
althought i just thought the question up you get the method: as im sure you cant get soln's for minus roots! LOL but do u understand i am not so gud at maths so sorry if its all rong!
5. (Original post by gemma.....)
say for eg u have a question saying:

"what are the minimum/max points on the graph 2x^2 + 8x - 5
you would do (x+4)(x+4) which gives you x^2 + 8x + 16 but we want -5 so we would take off 21 thus giving:
(x+4)^2 -21
this then makes
x= +/- the square root of -21 -4
althought i just thought the question up you get the method: as im sure you cant get soln's for minus roots! LOL but do u understand i am not so gud at maths so sorry if its all rong!
Yes I understand. Completing the square is straight forward, but I think you can get solutions for minus roots?!?! Im not sure. But thats the answer to 'urbanflavaz' question, gemmas response.
6. (Original post by urban_flavaz)
ok, my teacher was going on about how to do this by putting the equation in complete the square form and then using that 2 find out the vertex of the graph but i've kinda 4gotten how 2 do this and need some help. i'm not sure if this will come up or not but it'll probz be the last question. so how do u find out the vertex of a quadratic graph is basically wat i need an answer to. thanks
Complete the square of a quadratic.

Using:

y = f(x)

When: y = f(x + a) , graph shifted a places to the left.

When: y = f(x) + b , graph shifted b places up.

So: An example:

y = x² + 4x - 20 =0

y = (x + 2)² - 4 - 20 = 0

y = (x + 2)² - 24 = 0

Therefore:

The graph is y = x² to the left two places and down 24 places.

If you take the vertex of the parlabra (sp) to be (0,0) initially then it is (-2, -24) now.

I hope that helps.
7. Minus Square Roots are Imaginary Numbers.
8. Can anyone confirm / deconfirm what I have said above?
9. You complete the square to find the minimum point of a quadratic. So you will be finding (a) the minimum value for ax² + bx + c and (b) the value of x which gives this minimum value. If you draw a graph of y = ax² + bx + x you will therefore be finding the minimum value for y and the corresponding value for x which gives this. So you end up with two coordinates.

Completing the square gives:

ax² + bx + c
a(x² + (b/a)x + (c/a))
a[ (x + (b/2a))² - (b²/4a²) + (c/a) ]
a[ (x + (b/2a))² + (4ac - b²)/4a² ]
a[ (x + (b/2a))² ] + (4ac - b²)/4a

Now put the part inside the brackets equal to 0:

x + (b/2a) = 0
x = (-b/2a)

To get the value of x which given a min. value for (ax² + bx + c), then substitute this back into the equation to find what it gives:

a(-b/2a)² + b(-b/2a) + c
a(b²/4a²) - (b²/2a) + c
ab²/4a² - 2ab²/4a² + 4a²c/4a²
(ab² - 2ab² + 4a²c)/4a²
(4a²c - ab²)/4a²
(4ac - b²)/4a

Recognise this from anywhere? Have a look up if you don't - it's the part outside the square brackets. So we can conclude that by completing the square and equating the part inside the brackers containing x to zero, and finding a value for x (which gives a minimum), we don't need to substitute the whole thing back into the equation - just look at the part outside and that's your minimum point.

This might seem complicated for GCSE, but try it with numbers following it through - the a, b and c work just as well (but look more complicated), and mean that I can prove it for all values of these constants, and not just 4x² - 2x + 1.
10. (Original post by mik1a)
You complete the square to find the minimum point of a quadratic. So you will be finding (a) the minimum value for ax² + bx + c and (b) the value of x which gives this minimum value. If you draw a graph of y = ax² + bx + x you will therefore be finding the minimum value for y and the corresponding value for x which gives this. So you end up with two coordinates.

Completing the square gives:

ax² + bx + c
a(x² + (b/a)x + (c/a))
a[ (x + (b/2a))² - (b²/4a²) + (c/a) ]
a[ (x + (b/2a))² + (4ac - b²)/4a² ]
a[ (x + (b/2a))² ] + (4ac - b²)/4a

Now put the part inside the brackets equal to 0:

x + (b/2a) = 0
x = (-b/2a)

To get the value of x which given a min. value for (ax² + bx + c), then substitute this back into the equation to find what it gives:

a(-b/2a)² + b(-b/2a) + c
a(b²/4a²) - (b²/2a) + c
ab²/4a² - 2ab²/4a² + 4a²c/4a²
(ab² - 2ab² + 4a²c)/4a²
(4a²c - ab²)/4a²
(4ac - b²)/4a

Recognise this from anywhere? Have a look up if you don't - it's the part outside the square brackets. So we can conclude that by completing the square and equating the part inside the brackers containing x to zero, and finding a value for x (which gives a minimum), we don't need to substitute the whole thing back into the equation - just look at the part outside and that's your minimum point.

This might seem complicated for GCSE, but try it with numbers following it through - the a, b and c work just as well (but look more complicated), and mean that I can prove it for all values of these constants, and not just 4x² - 2x + 1.
I thought that was quite neat.. didn't expect it to work so well!
11. (Original post by mik1a)
I thought that was quite neat.. didn't expect it to work so well!
that was very good - u r a genius, give me ur brain for the maths exam on tuesday i dont grasp maths!

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