Zenmo
  • Zenmo
Eliminate the parameter to find a Cartesian equation of the curve for −7 ≤ y ≤ −3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Zenmo
  • Zenmo
x = 1 − t^2, y = t − 5, −2 ≤ t ≤ 2
triciaal
  • triciaal
|dw:1449468787649:dw|
Zenmo
  • Zenmo
(a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as t increases. I have already done that. Now for Eliminate the parameter to find a Cartesian equation of the curve for −7 ≤ y ≤ −3.

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Zenmo
  • Zenmo
|dw:1449469105670:dw|
triciaal
  • triciaal
|dw:1449468935540:dw|
Zenmo
  • Zenmo
What would the Cartesian equation be?
triciaal
  • triciaal
not sure working on it
Zenmo
  • Zenmo
the interval −7 ≤ y ≤ −3 makes me confuse, don't know what to do with that part.
triciaal
  • triciaal
@zepdrix
triciaal
  • triciaal
|dw:1449470130892:dw|
triciaal
  • triciaal
what is the equation for a horizontal parabola that opens left? and moved down 5 units
Zenmo
  • Zenmo
y = (x - h)^2 + k ?
zepdrix
  • zepdrix
Zen you made one little boo boo when you were constructing your equation.\[\large\rm x=1-t^2\]Solving for t gives us\[\large\rm t=\pm\sqrt{1-x}\] Taking the square root of a square gives you two roots! The positive and negative roots. So that's all you're missing from your Cartesian Equation,\[\large\rm y=\pm\sqrt{1-x}-5\]But ... we would like this written without the plus/minus, so we should solve for x in this equation.
Zenmo
  • Zenmo
@zepdrix how would I write this without the plus/minus to solve for x?
zepdrix
  • zepdrix
Ignore the plus/minus for a second. Pay attention to my last piece of advice, "solve for x". Do you know how to apply the steps to make that happen? :)
zepdrix
  • zepdrix
That involves moving everything to the left side of the equation.
iwanttogotostanford
  • iwanttogotostanford
@zepdrix um hello?>.. is till need help
Zenmo
  • Zenmo
|dw:1449472003536:dw|
iwanttogotostanford
  • iwanttogotostanford
@zepdrix im not sure if its 6 or -6 yet!!!
zepdrix
  • zepdrix
Ah close! :) You mixed up one of your steps though.
zepdrix
  • zepdrix
To undo the -5, you added 5 to each side, good. To undo `square root`, you do not want to apply `square root`.
iwanttogotostanford
  • iwanttogotostanford
@zepdrix hello!
Zenmo
  • Zenmo
\[y+5=\sqrt{1-x}\] then what?
zepdrix
  • zepdrix
Undo the `square root` by `squaring`! :)
zepdrix
  • zepdrix
The squaring will also `get rid of the plus/minus`, that's why we didn't need to worry about it when solving for x.
zepdrix
  • zepdrix
\[\large\rm (y+5)^2=\sqrt{1-x}~^2\]
iwanttogotostanford
  • iwanttogotostanford
@zepdrix it is 6, correct??
iwanttogotostanford
  • iwanttogotostanford
@zepdrix please just tell me that please
Zenmo
  • Zenmo
@zepdrix \[-y^2-10y+24=x?\]
zepdrix
  • zepdrix
I think it's better to `not` expand out the left side. Either way is fine though. If you're going to expand it out, be care with distributing the negative, cause I think you missed one on the 24.
zepdrix
  • zepdrix
\[\large\rm (y+5)^2=1-x\]\[\large\rm 1-(y+5)^2=x\]
zepdrix
  • zepdrix
That's a good place to stop right there ^ Because it's in the vertex form of a parabola.
Zenmo
  • Zenmo
what happen to the -x? the minus sign
Zenmo
  • Zenmo
would it be \[-(y+5)^2+1\]?
zepdrix
  • zepdrix
Sorry I did two steps in one there :) I can slow it down.\[\large\rm (y+5)^2=1-x\]Subtract 1 from each side,\[\large\rm (y+5)^2-1=-x\]Multiply each side by -1,\[\large\rm 1-(y+5)^2=x\]
Zenmo
  • Zenmo
@zepdrix the answer is wrong when I put \[1-(y+5)^2=x\] for "Eliminate the parameter to find a Cartesian equation of the curve −7 ≤ y ≤ −3"
zepdrix
  • zepdrix
Hmm :d
Zenmo
  • Zenmo
but it's fine, I ran out of attempts for this problem :/
zepdrix
  • zepdrix
Aw lame ;c
zepdrix
  • zepdrix
That should have been correct. Was your answer box set up like this maybe?\[\large\rm y=\fbox{Put answer here}\]
Zenmo
  • Zenmo
1 Attachment
zepdrix
  • zepdrix
You chose the wrong graph, could that maybe be why it was wrong? :o Hmmm Or the red x means that the equation was also wrong?
zepdrix
  • zepdrix
Blah >.< Oh well..
Zenmo
  • Zenmo
the correct graph is above the chosen one, but for some reason it marked the one below it.

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