Question

Eliminate the parameter.

x = t^2 + 3, y = t^2 - 2

Answer 1

@e.mccormick @satellite73

Answer 2

@agreene @Best_Mathematician @Christos @Danny_Boy @enya.gold @flutterflies

Answer 3

Solve for t, set themm both = to each other. Then you have it in x and y. That is probably what they mean, because t is generally the parameter.

Answer 4

the answer options are
y = x + 5, x ≥ 3

y = x2 - 5, x ≥ 3

y = x - 5, x ≥ 3

y = x2 + 5, x ≥ 3

Answer 5

OK. So they are having you unparamiterize it that way. This in Linear?

Answer 6

yes i think so

Answer 7

\[x = t^2 + 3\\
y = t^2 - 2\\ \therefore \\
\left[\matrix{
1 & 0 & | t^2 + 3\\
0 & 1 & | t^2 - 2
}\right]\Rightarrow
\left[\matrix{
1 & -t^2 & |\quad 3\\
-t^2& 1 & | - 2
}\right]\]

Answer 8

That would be one way of working towards it in Linear.... Which lets you see the relationships it started from. What you would be looking for as an answer is the solution space where that holds true.

Answer 9

Hmmm... Well, might not be quite the best matix to use to represent this. However, it does let me come up with an idea.

Answer 10

\(x-3=t^2\) And:
\(y+2=t^2\), but
\(t^2=t^2\), \(\therefore \, x-3 = y+2\)
Now to see how those relate to:
y = x + 5, x ≥ 3
y = x2 - 5, x ≥ 3
y = x - 5, x ≥ 3
y = x2 + 5, x ≥ 3

Answer 11

ok

Answer 12

Well, the \(x−3=y+2\Rightarrow x−1=y\) Ahh. I think I am losing answers because I did not square it. because I do not see that anywhere in the list.

Answer 13

I did a little search to see why I did not remember this in the linear I know. It is trig! Not linear. You in a trig class?

Answer 14

Or it can be...

Answer 15

im in pre calc we are doing trig right now

Answer 16

Kk. So this is most likely for trig. Now I see why I did not recognize the form right off. I took Trig and Alg, but skipped pre-calc. Probably something they did in there. I got some really ugly answers and some simple ones, but I did not get one that matched up with the possible choices, so I know I am doing something wrong.

@Mertsj @jim_thompson5910 Anyone care to dig out of this hole I found? I see the point of this, but using what I know does not give me any of the 4 answers listed, so I am obviously off somewhere.

Answer 17

why not isolate t^2 instead of isolate t?

Answer 18

that's much easier in my opinion

Answer 19

t^2=x-3
t^2=y+2
y+2=x-3
y=x-5

Answer 20

I tried that, but it did not get me where I wanted, so I must have made a mistake somewhere.

Answer 21

x = t^2 + 3, y = t^2 - 2

---------------

x = t^2 + 3

x-3 = t^2

t^2 = x-3

---------------

y = t^2 - 2

y = (x-3) - 2

y = x - 3 - 2

y = x - 5

Answer 22

AH HA! I made a simple algebra error. LOL! I got \(x-1\) and my brain got stuck there!

Answer 23

so it would be the third option?

Answer 24

Thanks guys!
@mgarcia634 OK... so if you do not make the stupid mistake I made... you get what Jim did. it is exactly the same thing Mertsj started and I did earlier when I did:
\[x-3 = t^2\\
y+2 = t^2\\
x-3 = y+2 \]
Then I went and added 2 rather than subtract it, which is where I confused the heck out of myself, and thereby confused you.

Answer 25

Yep! Third it is.

Answer 26

thanks guys!

Answer 27

Really sorry about the confusion.

Answer 28

it's fine