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#1
So I have a math assignment due tomorrow and I'm stuck at these questions.

Make the Graph and change to general Form for :

1)                                        2)

2/3(y + 5) = (x-4)²                            -1/2( y - 6) = (x + 3)²



These graph as a curve ( Parabola) or something I think.

If someone can give me a quick explanation on how to do these I'd be pretty grateful.

Thanks in advance,
ShaunDiel
Last edited by ShaunDiel at Feb 23, 2009,
#2
First distribute, then FOIL. You should be able to get on the right path.

AND THERE IS A MATHS THREAD.
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#4
Yes, they are parabolas, do you know what general form is?
Ax+By+C=0
um... that's all I got.
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#6
Distribute the 2/3 through the parenthesis. Then FOIL the (x-4)^2

Then you can put it into standard form.
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#7
Quote by Samdunhamss
Yes, they are parabolas, do you know what general form is?
Ax+By+C=0
um... that's all I got.


Well actually its: ax^2+bx+c=0

EDIT:arrange it in that form, then use this: x=-b±√b^2-4*a*c/2

I hope that makes sence.
Last edited by Just_Matt at Feb 23, 2009,
#8
solve (isolate) for y and then use graphing software on the internet.
i'll solve them right now, and if this closes ill pm you.

what is that exponent outside the brackets? a square?
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Last edited by tona_107 at Feb 23, 2009,
#9
Quote by AeroRocker
Distribute the 2/3 through the parenthesis. Then FOIL the (x-4)^2

Then you can put it into standard form.

Okay so I have

2/3y + 3.33 = (x - 4)²

for the first part right?
#10
for any graph, you want y = f(x) as a general form, where f(x) is a function involving x's.

in your case, you just need to multiply out the brackets (carefully!) and rearrange to said form.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
#11
Quote by ShaunDiel
Okay so I have

2/3y + 3.33 = (x - 4)²

for the first part right?


Well first of all you're gonna want to get into the habit of leaving improper fractions as-is. IE, leave it as 10/3.

Now you have to do (x-4)(x-4).
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#12
Quote by Sol9989
for any graph, you want y = f(x) as a general form, where f(x) is a function involving x's.

in your case, you just need to multiply out the brackets (carefully!) and rearrange to said form.


Actually, standard form for a quadratic (which is this) is Ax^2 + Bx + C = 0
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#13
for one i got

y = (3/2 (x)^2) - (12x) + (16)


i'll scan my work after if you want so you can learn maybe.
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Last edited by tona_107 at Feb 23, 2009,
#14
take off parantheses, the ()s, then put it this form: Y=ax^2 + bx + c
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#15
Quote by AeroRocker
Well first of all you're gonna want to get into the habit of leaving improper fractions as-is. IE, leave it as 10/3.

Now you have to do (x-4)(x-4).

My teacher told me that too

So, now I'm at.

2/3 y + 10/3 = x² -8x +16 


Or am I completely wrong.
Quote by tona_107
for one i got

y = (3/2 (x)^2) - (12x) + (16)


i'll scan my work after if you want so you can learn maybe.

Last edited by ShaunDiel at Feb 23, 2009,
#16
Ah, good old grade 9-10.
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#17
Quote by ShaunDiel
My teacher told me that too

So, now I'm at.

2/3 y + 10/3 = x² -8x +16


Or am I completely wrong.



Good! Now put it into standard quadratic form.

Isolate the y (subtract 10/3)
Continue to isolate y (divide both sides by 2/3, or multiply by 3/2)
Simplify.

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#19
and for number 2 i got

y = (-2x^2) - (12x) - (12)

don't include the brackets when you write it out, i'm just doing that to make it neater on the computer.
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#20
Quote by ShaunDiel
Advanced Math 12


lol Canda is stoooopid!!! Just kidding.

But that's actually weird. In 12th grade I'm taking BC Calculus.
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#21
Quote by AeroRocker
Actually, standard form for a quadratic (which is this) is Ax^2 + Bx + C = 0


i apologize for your inability to read the word "graph" which is what, at least part of, this problem is asking for.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
#22
ok im scanning it now... i'm pretty sure i'm right... i havn't done quadratics in about 5 years but i'm a university calc student so i would hope i'm right...

do you need help graphing as well?
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#23
no dude, you have to complete the square to get it in vertex form which is y=a(x-h)^2 + k. then the h and k is you vertex of your parabola and your a is how "fat" or "skinny" your parabola is.
Quote by tona_107
and for number 2 i got

y = (-2x^2) - (12x) - (12)

don't include the brackets when you write it out, i'm just doing that to make it neater on the computer.
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#24
Dude, this is math 12?

I'm assuming it's stretches, reflections and translations, amirite?

Anyhow, first begin by expanding both brackets to get rid of the brackets. Second, since this a standard quadratic relation (you did say it is a parabola), then isolate the y and put the rest on the other side. Last thing is to graph. Also, you could do it another way by putting it into standard quadratic form (it was posted above) and finding the square to get the vertex (also pointed out above).

That's what I remember from Math in grade 12, lol.
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#25
Quote by AeroRocker
lol Canda is stoooopid!!! Just kidding.

But that's actually weird. In 12th grade I'm taking BC Calculus.

Ya, I'm taking that first year cégép, but we only go up to grade 11 in Québec.
And Shaundiel:
You guys try to make it look all complicated and sophisamacated.
I'm that dude with the fro.
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Long story short, a whale flew out of the ocean, landed next to me and shot like a wall of water straight into my face.
#26
Quote by Sol9989
i apologize for your inability to read the word "graph" which is what, at least part of, this problem is asking for.


I accept your apology for my illogicality.





You are right, I missed that part.
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#27
Quote by AeroRocker
Good! Now put it into standard quadratic form.

Isolate the y (subtract 10/3)
Continue to isolate y (divide both sides by 2/3, or multiply by 3/2)
Simplify.


Hmm.
So I subtract 10/3 from the left side then subtract 10/3 from 16?

then Multiply Y by 3/2 and multiply 16 by 3/2?

And I'm only in 11th, I just took Math 12 So I could be free of math for next year
#28
Quote by RHCP987123
no dude, you have to complete the square to get it in vertex form which is y=a(x-h)^2 + k. then the h and k is you vertex of your parabola and your a is how "fat" or "skinny" your parabola is.



does it ask for it in vertex form or general form? i thought he said general form.
i don't even remember how to complete the square but it's a formula he should have no problem doing it himself.
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#29
Quote by ShaunDiel
Hmm.
So I subtract 10/3 from the left side then subtract 10/3 from 16?

then Multiply Y by 3/2 and multiply 16 by 3/2?

And I'm only in 11th, I just took Math 12 So I could be free of math for next year



Don't be stupid, take AP calculus. Unless the post-secondary institution you plan to apply for doesn't acknowledge it for credits and deems it a highschool course, then don't.
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#30
Quote by djszkoda
Don't be stupid, take AP calculus. Unless the post-secondary institution you plan to apply for doesn't acknowledge it for credits and deems it a highschool course, then don't.

Yeah, I shouldn't have said that actually. I plan on Taking Pre-Cal & Cal, I just didn't want too many hard courses in one semester.

I'm just starting to get into math as you can clearly see. Slacking off finally caught up with me.
#31
Quote by ShaunDiel
Hmm.
So I subtract 10/3 from the left side then subtract 10/3 from 16?

then Multiply Y by 3/2 and multiply 16 by 3/2?

And I'm only in 11th, I just took Math 12 So I could be free of math for next year


EVERYTHING you do to one side, you apply to EVERYTHING on the other. You multiply both entire sides of the equation by 3/2. This is true for any operation in an equation. for the first step, you technically subtracted 10/3 from the entire equation, but it simplifies by only being subtracted from the 16. get it?
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#32
Quote by ShaunDiel
Yeah, I shouldn't have said that actually. I plan on Taking Pre-Cal & Cal, I just didn't want too many hard courses in one semester.

I'm just starting to get into math as you can clearly see. Slacking off finally caught up with me.


I hated math, but I needed it for university. So I busted my ass. I remember getting a 97% GPA in the first semester of grade 12, but a 91% GPA in the second because I had no sciences and all sh!tty math/english courses. Not fun!
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#33
Quote by AeroRocker
EVERYTHING you do to one side, you apply to EVERYTHING on the other. You multiply both entire sides of the equation by 3/2. This is true for any operation in an equation. for the first step, you technically subtracted 10/3 from the entire equation, but it simplifies by only being subtracted from the 16. get it?

I apologize in advance for how retarded I am

So I have to multiply even my x^2 and -8x by 3/2?
#34
Yes sireee.
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#35
Yes, EVERYTHING.

That's the golden rule - what you do to one side must be done to the other side.
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#36
Quote by djszkoda
Yes, EVERYTHING.

That's the golden rule - what you do to one side must be done to the other side.



unless you're in physics....... hehe
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#38
Quote by ShaunDiel
Guys, I really appreciate the help.

So, now I have.

Y = 2/3 x^2 - 12x ?

I have a feeling this is way wrong.


You left out a number.
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#40
Quote by ShaunDiel
Advanced Math 12



in manitoba we started that in Advance grade 10 and now we are doing it more in standard pre calculus grade 11.
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