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# aight.... math geniuses come here O_O

Discussion in 'Off-Topic' started by Sober_Irishman69, Mar 16, 2004.

1. ### Sober_Irishman69IncGamers Member

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aight.... math geniuses come here O_O

yea, im a high school student still taking Alg II, so i need a lil help (parabolas)

our tests/quizzes always consist of a bonus, usually one i can look at and work out in my head. however, this time, i ran into a paticularly nasty one:

write a quadratic equation in the form y=a(x-h) [squared, the little 2, cant type the exponent] + k

the point are -1,10 0,6 and 2,** (<--- i typed correctly, yes)

ummm any help about methods to solve this beast? i dont get credit if i solve it like a week later, but knowin the mthod will help... thanks guys!

-Sober

o yea... i can tell you that i dont think any of the mentioned points are the vertex >_< ty!

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^ symbol means squared.

The form of the quardratic equation is:

y=a(x-h)^2 + k

right?

and then you just say...the point are -1,10 0,6 and 2,**

-1
10
0
6
2
**

right?

What are those points of?
Are they points that the parabola intersects or what?

If so, am I understanding it correctly that:

You have a parabola that intersects the following points [-1, 10, 0, 6, 2, **] and you need to write an equation for this parabola in the form of: y=a(x-h)^2 + k?

Is this the problem you are presenting to us? :spy:

edit: or do you mean the points are (X,Y), (-1,10), (0,6), (2,**) of the parabola?

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Ok....if my assumptions are correct...including my edit...then here is the answer.

Kinda

I feel guilty about giving you the answer, so Ill give it to you in a different form and let you play with it.

The equation of the parabola that passes thru the points (-1,10), (0,6), (2,**)
is:

y = 15x^2 + 11x + 6

There ya go, have fun :howdy:

4. ### StevinatorIncGamers Member

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I'm guessing we're supposed to figure out the values for h and k. if I recall, these values help determine the length and girth of the parabola--there's better math terms for such things...

If i remember correctly you should be substituting and solving.

solve your equation for either h or k. throw in the values for x and y. then you should get k=a bunch of stuff

the got back the the original equation and throw a bunch of stuff in where k was...solve that for h.

now you know what h is. go back to your original and solve that for k
that should do it....I think. I'm not mth genius...but i waded through it in college.

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Yea, thats kinda how to do it. Note: I did it with a computer that easily generated the proper equation....its a nifty program called Excel .

BUT, If you wanna do it by hand, this is the methodology. Basically you have 3 unknowns (a, h, k) and 3 equations. Your 3 equations are:

y=a(x-h)^2 + k
(-1,10), (0,6), (2,**)

10=a(-1-h)^2+k
6=a(0-h)^2+k
**=a(2-h)^2+k

The basic rule of thumb in math is you have to have as many equations as you do unknowns if you want to solve for the unknowns. So, here you have 3 unknowns and 3 equations, just solve an equation for one variable and substitute, etc...

6. ### Sober_Irishman69IncGamers Member

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you know, i actually tried that, i ummm.... kept getting crap like -4-2ah all the time... or some form of that. one time i actually thought i had the answer, but it turned out to be -1(4+2ah) same thing...

ty for the help so far though

7. ### Sober_Irishman69IncGamers Member

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sry for the dbl post guys >< im a lil excited on figuring this stuff out ^_^