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April 28, 2024, 10:45:45 am

Author Topic: some help for a yr 11 (:  (Read 7912 times)  Share 

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ben_ess12

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some help for a yr 11 (:
« on: February 12, 2011, 12:11:57 pm »
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sorry to fill this forum up with a unit 1/2 question but i needsome help :)

i have an equation to sketch a hyperbola or however you spell it,
i have found the y intercept, done the asymptotes? all the transformations and im just havin trouble gettin the x intercepts

the equation is

y=-2/(x-1)2 +3 (note that the +3 isnt part of the denominator in the fraction but next to, so it moves it 3 units left)

i just cant work out how to get the 2 x intercepts, the y intercept is 1 and the calculator gives me the x intercepts as
0.183 and 1.816
any help would be amazingg
thanks :)
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Re: some help for a yr 11 (:
« Reply #1 on: February 12, 2011, 12:18:46 pm »
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X-intercepts, y = 0
-2/(x-1)^2 + 3 = 0
-2/(x-1)^2 = -3
-2 = -3(x-1)^2
2/3 = (x-1)^2
(x-1)^2 = 2/3
(x-1) = sqrt(2/3) or -sqrt(2/3)
x = 1 + sqrt(2/3) pr 1 - sqrt(2/3)

Which gives the values ~1.816 and ~0.184 respectively.
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Re: some help for a yr 11 (:
« Reply #2 on: February 12, 2011, 12:20:13 pm »
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y=-2/(x-1)^2 +3
let y=0
0= -2/(x-1)^2 +3
-3 = -2/(x-1)^2
3 = 2/(x-1)^2
3((x-1)^2)=2
(x-1)^2= 2/3
x-1 = (+/-)sqrt(2/3)

x= sqrt(2/3) +1
x= 1 - sqrt(2/3)

Can someone check if I did that right.
Sorry I don't know how to use latex so yeah :)

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nacho

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Re: some help for a yr 11 (:
« Reply #3 on: February 12, 2011, 12:20:56 pm »
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well, to correct you, the graph moves three units up, not to the left :D

anyway, to find the x-int, put y = 0
thus you have:

0 = -2/(x-1)^2   + 3

-3 = -2/(x-1)^2
-3(x-1)^2 = -2
Note, at this point you can use the quadratic formula, but that's longer however, since it hasn't been applied in the prev two posts, here:
-3(x-1)^2 + 2 = 0

expand (x-1)^2
-3(x^2 - 2x + 1) +2 = 0
multiple by -3 and simplyife the equation:

-3x^2 + 6x - 1 = 0

use quadratice formula, (which i assume you're familiar with):
a= -3    b= 6   and c= -1
« Last Edit: February 12, 2011, 12:24:05 pm by nacho »
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Water

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Re: some help for a yr 11 (:
« Reply #4 on: February 12, 2011, 12:21:59 pm »
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Hey, its a truncus. If you manipulate it etc..



You should get + or - square root 2/3, then you bring the -1 over.

And your x intercepts should and 0.53 and 0.183...

That should give u your x intercepts, I don't see how it is wrong. Graph it on your cas calculator and check your gallary of graphs. Hyperbola is when u have 1/x




Your x = +1 and y = +3 is merely your asymptotes :)
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ben_ess12

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Re: some help for a yr 11 (:
« Reply #5 on: February 12, 2011, 12:23:09 pm »
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thankyou thankyopu thankyou! :) :)
and truncus thats what i mean ;) too many new graphs in first week of school to remember :)
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Re: some help for a yr 11 (:
« Reply #6 on: February 12, 2011, 12:24:56 pm »
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A truncus is when the degree is '-n' and 'n' is a even natural number

A hyperbola is when the degree is '-n' and 'n' is an odd natural number


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Re: some help for a yr 11 (:
« Reply #7 on: February 12, 2011, 12:27:52 pm »
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Wtf....................

........up2 graphs like that already?

I'm still learning m = tan theta -.-

ben_ess12

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Re: some help for a yr 11 (:
« Reply #8 on: February 12, 2011, 12:33:46 pm »
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@nacho
thanks for the quad formula as well, i was thining of using that because i remember in yr10 thats another way to use it, but i also wanted to see the other as that is what my teacher will be wanting i think...

@Rohitpi, sorry but i actually have no idea what you mean :)
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pi

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Re: some help for a yr 11 (:
« Reply #9 on: February 12, 2011, 12:35:27 pm »
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@Rohitpi, sorry but i actually have no idea what you mean :)

By 'degree' I mean power. The power of your truncus was -2 as if the denominator has a power of '2' and as it was in the denominator it would be '-2'.

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Re: some help for a yr 11 (:
« Reply #10 on: February 12, 2011, 01:15:58 pm »
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@nacho
thanks for the quad formula as well, i was thining of using that because i remember in yr10 thats another way to use it, but i also wanted to see the other as that is what my teacher will be wanting i think...

@Rohitpi, sorry but i actually have no idea what you mean :)

He means to say that functions with an EVEN power, i.e. 2, 4, 6 etc etc or -2, -4, -6 etc take the shape of a truncus, (for example 1/x^2 is the basic truncus) whilst a hyperbola is the same except with ODD 'powers' (e.g. 1/x)

Hope that makes sense!
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luffy

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Re: some help for a yr 11 (:
« Reply #11 on: February 12, 2011, 04:32:28 pm »
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@nacho
thanks for the quad formula as well, i was thining of using that because i remember in yr10 thats another way to use it, but i also wanted to see the other as that is what my teacher will be wanting i think...

@Rohitpi, sorry but i actually have no idea what you mean :)

He means to say that functions with an EVEN power, i.e. 2, 4, 6 etc etc or -2, -4, -6 etc take the shape of a truncus, (for example 1/x^2 is the basic truncus) whilst a hyperbola is the same except with ODD 'powers' (e.g. 1/x)

Hope that makes sense!

Yeah, however, polynomials with degree 2,4,6 etc would not be a truncus when graphed. :P

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Re: some help for a yr 11 (:
« Reply #12 on: February 12, 2011, 05:06:04 pm »
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@nacho
thanks for the quad formula as well, i was thining of using that because i remember in yr10 thats another way to use it, but i also wanted to see the other as that is what my teacher will be wanting i think...

@Rohitpi, sorry but i actually have no idea what you mean :)

He means to say that functions with an EVEN power, i.e. 2, 4, 6 etc etc or -2, -4, -6 etc take the shape of a truncus, (for example 1/x^2 is the basic truncus) whilst a hyperbola is the same except with ODD 'powers' (e.g. 1/x)

Hope that makes sense!

Yeah, however, polynomials with degree 2,4,6 etc would not be a truncus when graphed. :P

It does take the general shape though.
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luffy

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Re: some help for a yr 11 (:
« Reply #13 on: February 12, 2011, 05:32:00 pm »
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@nacho
thanks for the quad formula as well, i was thining of using that because i remember in yr10 thats another way to use it, but i also wanted to see the other as that is what my teacher will be wanting i think...

@Rohitpi, sorry but i actually have no idea what you mean :)

He means to say that functions with an EVEN power, i.e. 2, 4, 6 etc etc or -2, -4, -6 etc take the shape of a truncus, (for example 1/x^2 is the basic truncus) whilst a hyperbola is the same except with ODD 'powers' (e.g. 1/x)

Hope that makes sense!

Yeah, however, polynomials with degree 2,4,6 etc would not be a truncus when graphed. :P

It does take the general shape though.

The general shape of a truncus? What do you mean?

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Re: some help for a yr 11 (:
« Reply #14 on: February 12, 2011, 07:04:35 pm »
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@nacho
thanks for the quad formula as well, i was thining of using that because i remember in yr10 thats another way to use it, but i also wanted to see the other as that is what my teacher will be wanting i think...

@Rohitpi, sorry but i actually have no idea what you mean :)

He means to say that functions with an EVEN power, i.e. 2, 4, 6 etc etc or -2, -4, -6 etc take the shape of a truncus, (for example 1/x^2 is the basic truncus) whilst a hyperbola is the same except with ODD 'powers' (e.g. 1/x)

Hope that makes sense!

Yeah, however, polynomials with degree 2,4,6 etc would not be a truncus when graphed. :P

Oops, I meant like 1/x^2 and stuff so technically -2 etc
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