GCSE maths - vertex of graphs

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

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!

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

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!