Question

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

Answer 1

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

Answer 2

can you explain some more?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

\(x-4 = \frac{2}{3} t\)

\(\frac{2}{3}t = x-4\)

\(t = \frac{3}{2} x - 6\)

Answer 10

@whovianchick no, it is not a function

but vinny has some work to do before we get there

Answer 11

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.

Answer 12

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

Answer 13

not quite

Answer 14

sign on the 3??

Answer 15

positve

Answer 16

and when you have done that we are told that \(y^2 = t^2\)

so stuff your new equation into that

Answer 17

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

Answer 18

thanks whovian:-)

Answer 19

keep going vinny, we are nearly done!

Answer 20

y=3x=3

Answer 21

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

Answer 22

ah they are squared gh

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

yeah thats what i did

Answer 27

thanks guys

Answer 28

no vinny

it is NOT a function

Answer 29

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

Answer 30

you will see why if we finish this properly

but if you are really tired, go to bed

Answer 31

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

Answer 32

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

Answer 33

"thatis ture"
that is true

Answer 34

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

Answer 35

@vinny0515
are you done for the day?

Answer 36

this is what we have here

Answer 37

No it's not, and both are functions

Answer 38

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

Answer 39

no it is not

Answer 40

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

Answer 41

y=3x+3 is the answer

Answer 42

how do i answer it then

Answer 43

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

Answer 44

\(y^2=(3x+3)^2\) 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

Answer 45

irsih boy is right

Answer 46

@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

Answer 47

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