#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 :

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

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.

AND THERE IS A MATHS THREAD.

#3

Yeah but no one really looks at it.. Just a few people and they're offline. I checked.

#4

Yes, they are parabolas, do you know what general form is?

Ax+By+C=0

um... that's all I got.

Ax+By+C=0

um... that's all I got.

#5

Yeah for the First one my sister helped me out a bit and gave me this:

But she's gone out now so I can't ask her how she got it or if it's right.

`Y = 3x² - 24x +53`

But she's gone out now so I can't ask her how she got it or if it's right.

#6

Distribute the 2/3 through the parenthesis. Then FOIL the (x-4)^2

Then you can put it into standard form.

Then you can put it into standard form.

#7

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?

i'll solve them right now, and if this closes ill pm you.

what is that exponent outside the brackets? a square?

*Last edited by tona_107 at Feb 23, 2009,*

#9

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.

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

#11

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).

#12

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

#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.

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

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

*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

#15

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.

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.

#17

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.

#18

Ah, good old grade 9-10.

Advanced Math 12

*Last edited by ShaunDiel at Feb 23, 2009,*

#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.

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.

#20

Advanced Math 12

lol Canda is stoooopid!!! Just kidding.

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

#21

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.

#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?

do you need help graphing as well?

#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.

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.

#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.

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.

#25

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.

#26

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.

#27

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

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.

#29

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.

#30

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

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?

#32

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!

#33

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.

#35

Yes, EVERYTHING.

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

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

#36

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

#37

Guys, I really appreciate the help.

So, now I have.

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

I have a feeling this is way wrong.

So, now I have.

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

I have a feeling this is way wrong.

#38

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.

#39

You left out a number.

My bad

Y = 2/3 x^2 - 12x + 18.9?

#40

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.