find the direct relationship between x and y and determine whether the parametric equations determine y as a function of x
x=(1/3)t-1 y^2=t^2

- anonymous

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- IrishBoy123

get t in terms of x from this one: x=(1/3)t-1

- anonymous

can you explain some more?

- IrishBoy123

you have x=(1/3)t-1, ie an equation giving x in terms of t
work it so that you have t in terms of x, ie t = ....
use normal algebra

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## More answers

- anonymous

is this correct so far? \[t=((1/3)-1)/x\]

- whovianchick

y=3x+3, yes it's a function

- IrishBoy123

isn't it \[x=\frac{1}{3}t-1\]???
all you have to do is move the t across to the RHS and the x across to the RHS, applying usual rules of maipulation

- anonymous

oh so just t= \[t=\frac{ 1 }{3 }x-1\]

- whovianchick

Well, you have to find out if y is a function of x. first of all, if x=1/3t-1, then 3x=t-3, and t=3x+3. Then if y^2=t^2, y=t, so y=3x+3. Which is a function. I don't know where you got t=1/3x-1. That's not right

- IrishBoy123

no, i said "applying usual rules of manipulation". you can't just lift them out(!!) and switch them around
an example
if \(x = \frac{2}{3} t + 4\)
then
\(x-4 = \frac{2}{3} t\)
\(\frac{2}{3}t = x-4\)
\(t = \frac{3}{2} x - 6\)

- IrishBoy123

@whovianchick no, it is not a function
but vinny has some work to do before we get there

- whovianchick

Except that your math is wrong and you're forgetting y. Recheck your math. You can't just switch t and x, you have to solve for t. It's called 7th grade math. I'm in 11th grade math.

- anonymous

sorry his wording confused me its t=3x-3 correct?

- IrishBoy123

not quite

- IrishBoy123

sign on the 3??

- anonymous

positve

- IrishBoy123

and when you have done that we are told that \(y^2 = t^2\)
so stuff your new equation into that

- anonymous

ugh sorry im working on little sleep and silly mistakes are getting me'

- whovianchick

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- IrishBoy123

thanks whovian:-)

- IrishBoy123

keep going vinny, we are nearly done!

- anonymous

y=3x=3

- whovianchick

You're welcome. You had him working with the wrong equation, and I'm nitpicky like that

- anonymous

ah they are squared gh

- IrishBoy123

yes so just square them, DO NOT multiply/expand out the x side,...!

- IrishBoy123

"You had him working with the wrong equation,"
when?

- whovianchick

if y^2 = t^2, then y=t Replace y for t in the equation: y=3x+3, which is a function

- anonymous

yeah thats what i did

- anonymous

thanks guys

- IrishBoy123

no vinny
it is NOT a function

- whovianchick

No prob. Sorry @IrishBoy123 earlier he said something like 1/3x-1 = t and I thought it was you

- IrishBoy123

you will see why if we finish this properly
but if you are really tired, go to bed

- whovianchick

@IrishBoy123 UMMMMMMMMMMMM y=3x+3 IS a function

- IrishBoy123

it is a function, thatis ture, but that is not what we have here

- IrishBoy123

"thatis ture"
that is true

- whovianchick

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- whovianchick

That's the answer, I'm confused what you're saying

- IrishBoy123

@vinny0515
are you done for the day?

- IrishBoy123

this is what we have here
|dw:1442362770822:dw|

- whovianchick

No it's not, and both are functions

- anonymous

so the answer for the quesrion is y=3x+3

- IrishBoy123

no it is not

- whovianchick

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- whovianchick

What is the answer, then? We've replaced y for t, and we're done. What the heck are you talking about?

- whovianchick

y=3x+3 is the answer

- anonymous

how do i answer it then

- whovianchick

The way I told you to. y=3x+3, and yes it's a function

- IrishBoy123

\(y^2=(3x+3)^2\) is what you have
if you want to simplify it you can say \(y= \color{red}\pm (3x+3)\)
but whatever you do, x maps to 2 y values
it is not a function

- anonymous

irsih boy is right

- whovianchick

@IrishBoy123 I understand that. But it's saying as one line, is every solution a function. The answer is yes. Both of those equations are functions

- IrishBoy123

it is not a function as you can get 2 y's for every x
y = 3x+3 is a function
y = -(3x+3) is a function
this is both and is not a function
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- IrishBoy123

and that's us done
thank you both:p

- anonymous

so i write the answer as y is not a function

- whovianchick

Y is a function. I think you're forgetting the concept of a function, @IrishBoy123 , it's ok if there are multiple y's for every x. What you don't want is multiply x's for every y. THAT is the definition of a function, and even with your skewed logic, it's STILL a function. Stop confusing @vinny0515. YOU ARE WRONG

- IrishBoy123

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