#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...?

talk to the teacher and use the book...?

#3

what form does it need to be in? standard form or vertex for or what?

#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.

legitimate question by the way.

#5

America - math

Elsewhere - maths

Because maths is short for mathematics.

Elsewhere - maths

Because maths is short for mathematics.

#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

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.

#8

I like parabolas. I also like hyperbolas.

But I excel most in hyperbole.

But I excel most in hyperbole.

#9

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

I like parabolas. I also like hyperbolas.

But I excel most in hyperbole.

#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.

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

#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

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

#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

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

#15

I like parabolas. I also like hyperbolas.

But I excel most in hyperbole.

Hyperboles are the best motherf

**ucking thing in the WHOLE UNIVERSE!**

#16

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.

#17

Is wills nots helps yous withs thes maths. Is wills bangs olds ladies insteads.

Hooray for ignorance.

#18

Hooray for ignorance.

Yous seems likes you needs a chills pills.