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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Cappuccinos on January 20, 2011, 04:14:10 pm

Title: Find the equation...?
Post by: Cappuccinos on January 20, 2011, 04:14:10 pm: So I'm supposed to find the equation for this graph
*The orange bit in the attached picture*
Its a modulus/absolute value function

I have no idea what to do..
All I've got so far is $y= -|f(x)|+1$ (which was obtained by entering random stuff into the calc until I got a similar looking graph :buck2:)

What steps are you supposed to do when solving type of question?
What was the original function? I also need to state the domain? $36<x<40$ right?
Sorry about my noobness :-\

Edit: Oh and the x-intercepts are at 36 & 40! NOT 35 & 40!
Title: Re: Find the equation...?
Post by: Romperait on January 20, 2011, 04:27:36 pm: Since you know two points, $(36, 0)$ $(38, 1)$ , then you can find gradient...

$\frac{1-0}{38-36}=\frac{1}{2}$

$y-1=\frac{1}{2}(x-38)$
$y=\frac{1}{2}x-18$

Since the modulus occurs at $(38, 1)$ , you can write the modulus function as $y=-|\frac{1}{2}x-19|+1$ .

Sorry if it's wrong / hard to understand. First time doing maths on a computer. :P

Edit: fix'd

Also $x\epsilon[36, 40]$

Edit 2: Run over rise tends to not work in these situations. =(
Title: Re: Find the equation...?
Post by: onur369 on January 20, 2011, 04:32:29 pm: You used the two-mean method if I am not wrong...

y-y1= m(x-x1)
Title: Re: Find the equation...?
Post by: Mao on January 20, 2011, 04:57:12 pm: $y = -\frac{1}{2}|x-38| +1$ is the equation you seek.
Title: Re: Find the equation...?
Post by: Cappuccinos on January 20, 2011, 05:03:24 pm: Sorry, I think I left the edit a bit late. Btw, I didn't mean to yell D:

Quote from: onur369 on January 20, 2011, 04:32:29 pm
You used the two-mean method if I am not wrong...

y-y1= m(x-x1)

Whats a two - mean method? Ah I shouldn't be doing methods should I? :/

Quote from: Romperait on January 20, 2011, 04:27:36 pm
Since you know two points, $(36, 0)$ $(38, 1)$ , then you can find gradient...

$\frac{38-35}{1-0}=2$

$y-1=2(x-38)$
$y=2x-75$

Since the modulus occurs at $(38, 1)$ , you can write the modulus function as $y=-|2x-76|+1$ .

Sorry if it's wrong / hard to understand. First time doing maths on a computer. :P

Edit: fix'd

Also $x\epsilon[36, 40]$

Hahah thanks! I get your jist. The answer wasn't correct though it's $y = -|\frac{1}{2}x-19| +1$ your theory was though :) Your error was $\frac{38-35}{1-0}=2$ it should be $\frac{1-0}{38-35}=$ $\frac{1}{2}$
Title: Re: Find the equation...?
Post by: Romperait on January 20, 2011, 05:08:44 pm: Quote from: Steph243 on January 20, 2011, 05:03:24 pm
Hahah thanks! I get your jist. The answer wasn't correct though it's $y = -|\frac{1}{2}x-19| +1$ your theory was though :) Your error was $\frac{38-35}{1-0}=2$ it should be $\frac{1-0}{38-35}=$ $\frac{1}{2}$

:-\

Ah well.

Thanks Mao, btw.
Title: Re: Find the equation...?
Post by: Cappuccinos on January 20, 2011, 05:09:27 pm: Haha actually Romperait you are right. I missed your newest edit :/
Title: Re: Find the equation...?
Post by: taiga on January 20, 2011, 09:15:25 pm: I didn't read much, but how does 38-35 = 2

:X
Title: Re: Find the equation...?
Post by: Pixon on January 20, 2011, 09:21:11 pm: OP originally said 35 instead of 36.
BTW..Romperait you noob. jks? :P
Title: Re: Find the equation...?
Post by: Romperait on January 20, 2011, 09:25:51 pm: Quote from: Pixon on January 20, 2011, 09:21:11 pm
OP originally said 35 instead of 36.
BTW..Romperait you noob. jks? :P

^-^