Help with math

2006-12-14 12:20:37

#1

Rec(+)n: Teh Hawtsawce; +4|6893

uh yeah, i need help with the maths. I was absent when we went over this stuff, and I'm not sure exactly how to do it, so here goes.

i have to identify the vertex (h,k) of the graph y = f(x), solve for f(x) by using f(x) = 0, and graph it labelling the vertex and intercepts, and finally use the graph to solve the inequality f(x)<0

here are the problems

1. f(x) = x^2 - 7x + 12

2. f(x) 2x^2 + 6x + 5

3. f(x) = -3x^2 + 7x - 4

i got the vertex and x, but i dont know how to graph it or solve the inequality

also, a miscelanous problem : i have to solve for x : log(x+1) + log(x-1) = 2

ill give karma to whoever helps, and thanks

Last edited by Rec(+)n (2006-12-14 12:22:45)

2006-12-14 12:28:11

#2

confused: Member; +10|6829|British Columbia

To graph simply choose a selection of values for x, calculate y and plot. To solve the inequality you must determine what part of the graph (above/below/left/right of the line has valid solutions) and shade the area (or otherwise mark it as the solution set). If the line is part of the solution set it is solid. If the line is not part of the solution set it should be hatched.

2006-12-14 12:33:07

#3

Rec(+)n: Teh Hawtsawce; +4|6893

alright, thanks alot man that helped a bit, any idea how to get x in the log problem?

2006-12-14 12:37:06

#4

Pernicious544: Zee Tank Skank; +80|7135|MoVal So-Cal

damn Im happy I dont have complex math anymore

2006-12-14 12:38:40

#5

Sydney: 2λчиэλ; +783|7278|Reykjavík, Iceland.

That's almost the same thing I'm learning right now in advanced math class, what grade are you in?

2006-12-14 12:39:19

#6

Rec(+)n: Teh Hawtsawce; +4|6893

PBAsydney wrote:
That's almost the same thing I'm learning right now in advanced math class, what grade are you in?

i am a freshman in college x.X

2006-12-14 12:43:27

#7

Sydney: 2λчиэλ; +783|7278|Reykjavík, Iceland.

Rec(+)n wrote:
PBAsydney wrote:
That's almost the same thing I'm learning right now in advanced math class, what grade are you in?
i am a freshman in college x.X

10th grade here , I'm 15.

Last grid graph assignment I was doing was to find the formula of the skew number (lol, no idea what it's called, something about f´(x) and the formula y-r = a(x-s) where (r,s) is the coordinates on the graph, and a is this so called skew number)

Complicated stuff indeed.

2006-12-14 12:43:36

#8

Paco_the_Insane: Phorum Phantom; +244|7079|Ohio

the log one is
log((x+1)(x-1))=2
log((x+1)(x-1))=log(100)
(x+1)(x-1)=(100)
x^2-1=100
x=sqrt.(101)

i think, still abit foggy in my memory

2006-12-14 12:47:14

#9

2tuff: Positive Karma Here!; +357|7210

ummmm 1+1=2....That i know for sure...

f(x) = x^2 - 7x + 12...Its still a bit unclear in that area. lol

Tuff

2006-12-14 12:51:08

#10

jimmanycricket: EBC Member; +56|7090|Cambridge, England

""also, a miscelanous problem : i have to solve for x : log(x+1) + log(x-1) = 2 ""

OK, i think this rearanges to this log((x+1)(X-1)) = 2

=> (X+1)(X-1)=100

=> X^2 -1 - 100 =0

=> X^2 -101 = 0

=> X= +/- sqrt(101)

==::EDIT::==

All you need to remember is that

Log(x) + Log(Y) = Log(X*Y)

and

Log(x) - Log(Y) = Log(X/Y)

Last edited by jimmanycricket (2006-12-14 12:55:30)

2006-12-14 12:51:46

#11

Sydney: 2λчиэλ; +783|7278|Reykjavík, Iceland.

f(x) = x^2 - 7x + 12

differed makes

f(x) = 2x - 7

put in 0

f´(0) = 2 * 0 - 7

f´(0) = -7

Has that something to do with it?

Edit: Looking at the above post, nope, I'm talking gibberish, I haven't learned anything about logs yet

Last edited by PBAsydney (2006-12-14 12:52:54)

2006-12-14 12:52:16

#12

Yaocelotl: :D; +221|7084|Keyboard

Rec(+)n wrote:
uh yeah, i need help with the maths. I was absent when we went over this stuff, and I'm not sure exactly how to do it, so here goes.

i have to identify the vertex (h,k) of the graph y = f(x), solve for f(x) by using f(x) = 0, and graph it labelling the vertex and intercepts, and finally use the graph to solve the inequality f(x)<0

here are the problems

1. f(x) = x^2 - 7x + 12

2. f(x) 2x^2 + 6x + 5

3. f(x) = -3x^2 + 7x - 4

i got the vertex and x, but i dont know how to graph it or solve the inequality

also, a miscelanous problem : i have to solve for x : log(x+1) + log(x-1) = 2

ill give karma to whoever helps, and thanks

You can solve that by using intregrals and determinants, nothing tuff. Case closed.

2006-12-14 12:54:16

#13

Rec(+)n: Teh Hawtsawce; +4|6893

yeah i got it

thanks alot guys

2006-12-14 12:55:39

#14

Paco_the_Insane: Phorum Phantom; +244|7079|Ohio

jimmanycricket wrote:
also, a miscelanous problem : i have to solve for x : log(x+1) + log(x-1) = 2

OK, i think this rearanges to this log((x+1)(X-1)) = 2

=> (X+1)(X-1)=100

=> X^2 -1 - 100 =0

=> X^2 -101 = 0

=> X= +/- sqrt(101)

==::EDIT::==

All you need to remember is that

Log(x) + Log(Y) = Log(X*Y)

and

Log(x) - Log(Y) = Log(X/Y)

ya, thats the formula, but for the answer, it cant be negative, because you cant take the log of a negative.
so always remember to check

2006-12-14 13:01:36

#15

Longbow: Member; +163|7081|Odessa, Ukraine

lmao ... the future of education will be like - maths forum , languages forum , etc . Lol

2006-12-14 13:05:21

#16

Sydney: 2λчиэλ; +783|7278|Reykjavík, Iceland.

Longbow wrote:
lmao ... the future of education will be like - maths forum , languages forum , etc . Lol

w00t! We can grow even fatter if we don't have to walk to school anymore!

2006-12-14 15:36:06

#17

Krauser98: Extra Green Please!; +53|7264|USA! USA! USA!

Can you guys write a paper for me while you're at it? I need it to be about 20 pages long, with works cited, single spaced. The topic is "How to prevent and solve the degredation of university lawns." It's due on Monday. I'll check back. I'll pay you guys in karma.

2006-12-14 15:39:20

#18

MorbiD.ShoT: Stormin' through the party; +322|7031

PBAsydney wrote:
f(x) = x^2 - 7x + 12

differed makes

f(x) = 2x - 7

put in 0

f´(0) = 2 * 0 - 7

f´(0) = -7

Has that something to do with it?

Edit: Looking at the above post, nope, I'm talking gibberish, I haven't learned anything about logs yet

Umm. This is not Calculus, this is Algebra II.

Absolutely nothing to do with what he wants to find out.

2006-12-14 15:40:13

#19

Ryan: Member; +1,230|7278|Alberta, Canada

There really should be a homework section on here! There's so many math questions lol

2006-12-14 15:42:24

#20

MorbiD.ShoT: Stormin' through the party; +322|7031

ryan_14 wrote:
There really should be a homework section on here! There's so many math questions lol

There is, right here http://forums.bf2s.com/viewtopic.php?id=45651

2006-12-14 15:44:06

#21

Ryan: Member; +1,230|7278|Alberta, Canada

MorbiD.ShoT wrote:
ryan_14 wrote:
There really should be a homework section on here! There's so many math questions lol
There is, right here http://forums.bf2s.com/viewtopic.php?id=45651

Ooh. I was thinking more of a subforum, but thats a little too big for this kind of thing i guess.

2006-12-14 15:47:11

#22

Cheez: Herman is a warmaphrodite; +1,027|6873|King Of The Islands

Mathematics << Plural
Math..........s

Say it. MAAAAATHSSS. Very good!

/"Excellent" Dinosaur Sticker

My state was founded by Batman. Your opinion is invalid.

2006-12-14 15:57:41

#23

some_random_panda: Flamesuit essential; +454|6825

confused wrote:
To graph simply choose a selection of values for x, calculate y and plot. To solve the inequality you must determine what part of the graph (above/below/left/right of the line has valid solutions) and shade the area (or otherwise mark it as the solution set). If the line is part of the solution set it is solid. If the line is not part of the solution set it should be hatched.

1. f(x) = x^2 - 7x + 12
0 = x^2 - 7x + 12
0 = (x-4)(x-3)
Therefore x-intercepts are 4 and 3, using the Null factor law.
The graph is a porabola, seeing as it is a quadratic equation.
Thus the vertex or turning point is halfway between the x-intercepts.
Sketch the curve (shaped like a U making contact with the x-axis at 4 and 3) and label the vertex or turning point.
^^^This is faster than plotting and saves a lot of time. You don't even need graph paper.

Finally, look for all values of y less than 0 and note the x-values.
The answer is then, depending on the graph:
{x: ?>x>?}
---OR---
{x: ?>x}U{x: ?<x} <------I think
(Yes, U is a mathematical symbol)

Just clarifying. DAMN I hate Quartic functions.

Last edited by some_random_panda (2006-12-14 16:15:50)

BF2S Forums Help with math

PBAsydney wrote:

Rec(+)n wrote:

PBAsydney wrote:

Rec(+)n wrote:

jimmanycricket wrote:

Longbow wrote:

PBAsydney wrote:

ryan_14 wrote:

MorbiD.ShoT wrote:

ryan_14 wrote:

confused wrote:

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