#1
I worked out how to solve the maths problem.
Last edited by kilbie at Aug 11, 2008,
#2
If you have the book and wanna learn it....


talk to the teacher and use the book...?
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#3
what form does it need to be in? standard form or vertex for or what?
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#4
why do people seem to always type it out as "maths"? why the 's'? is it a dialect thing?


legitimate question by the way.
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#5
America - math

Elsewhere - maths

Because maths is short for mathematics.
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#6
It'll be something like:

Y = -1X^2 -1

I know how to do these I just can't really be arsed at the moment :P
#7
Is wills nots helps yous withs thes maths. Is wills bangs olds ladies insteads.
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#8
I like parabolas. I also like hyperbolas.

But I excel most in hyperbole.
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#9
Quote by UncleCthulhu
It'll be something like:

Y = -1X^2 -1

I know how to do these I just can't really be arsed at the moment :P


I want the answer in this form. And I want to know HOW to get the answer.
#10
Quote by darkstar2466
I like parabolas. I also like hyperbolas.

But I excel most in hyperbole.


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#11
Haha I remember having to do this stuff a few years ago, it could be pretty annyoing.

Find out what form they are asking you to put it in (ex: standard), then plug the info you have into the basic formula . Take one point on the graph, then use it to measure off another point on the graph (ex: if its curved opposite of a normal parabola its -1), which will help you find the numbers for the remaining portion of the formula.

That was a crappy explanation and might not be right, but hey, I tried.
#12
y=(x-1)^2-1
And what is more, there's been a bloody purple nose and some bloody purple clothes that were messing up the lobby floor. It's just apartment house rules so all you 'partment fools remember : one man's ceiling is another man's floor.
#13
first way:

that's quadratic standard form.

y = ax^2 + bx + c

the formula of x coordinate of the vertex is = -b/2a

c is 0 in this case since it's given that y intercept is 0 so we have y = ax^2 + bx

and x coordinate of the vertex is 1

so 1 = -b/2a

therefore 2a = -b

when x = 1 y = -1 (the vertex)

- 1 = a (1)^2 + b(1)

= a - 2a
a = 1
b = - 2

the equation is y = x^2 - 2x
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#14
i remember doing this in 7th grade. are you in 7th grade?

way 2:

since the bases or x intercept are 0 and 2 the equation is: y = (x - 0) (x - 2)

y = x (x-2)
y = x^2 - 2x
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#15
Quote by darkstar2466
I like parabolas. I also like hyperbolas.

But I excel most in hyperbole.



Hyperboles are the best motherfucking thing in the WHOLE UNIVERSE!
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#16
Quote by rx_eb
i remember doing this in 7th grade. are you in 7th grade?

way 2:

since the bases or x intercept are 0 and 2 the equation is: y = (x - 0) (x - 2)

y = x (x-2)
y = x^2 - 2x


I just used the standard y=x^2, then moved it one to the right x -> x-1 then one down -1. y=(x-1)^2-1, which is exactly the same as your formula, just a different notation and theory.
And what is more, there's been a bloody purple nose and some bloody purple clothes that were messing up the lobby floor. It's just apartment house rules so all you 'partment fools remember : one man's ceiling is another man's floor.
#17
Quote by Munchlaxatives
Is wills nots helps yous withs thes maths. Is wills bangs olds ladies insteads.

Hooray for ignorance.
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#18
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Hooray for ignorance.


Yous seems likes you needs a chills pills.
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