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Skittles456

  • 3 years ago

@timo86m Is this B? http://prntscr.com/12c0cm

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  1. Skittles456
    • 3 years ago
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    http://prntscr.com/12c0hj - thats D

  2. timo86m
    • 3 years ago
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    I am looking at it :)

  3. Skittles456
    • 3 years ago
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    Take your time :)

  4. agent0smith
    • 3 years ago
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    For the first, since x=2t, then t=x/2. Replace t in the y=t^2+t+3 equation with x/2.. but you don't even need to do that. Can you find the y intercept of y=t^2+t+3? That's all you need to choose the answer. (what is y equal to when t=0? and note that when t=0, x=0 since x=2t)

  5. agent0smith
    • 3 years ago
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    First one isn't B.

  6. Skittles456
    • 3 years ago
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    Opps! I ment D not B

  7. Skittles456
    • 3 years ago
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    Let me double check to make sure D is my final answere lol

  8. timo86m
    • 3 years ago
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    http://graphsketch.com/parametric.php :D ooh

  9. agent0smith
    • 3 years ago
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    y-intercept is when x=0

  10. Skittles456
    • 3 years ago
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    well. I know its C now but y-intercept when x=0? what do you mean?

  11. agent0smith
    • 3 years ago
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    The y-intercept is... when x=0. Did you find the y-intercept of y=t^2+t+3? To find it: Replace t in y=t^2+t+3 with x/2, since t=x/2 \[\Large y=\left( \frac{ x }{ 2 } \right)^2+\frac{ x }{2 }+3 \] Two equations have the required y-intercept, so you'll have to plug in a couple of points to check which is correct.

  12. agent0smith
    • 3 years ago
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    This would be easier if the graphs weren't so tiny...

  13. Skittles456
    • 3 years ago
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    I can send them bigger lol. I have to run for a little. I hope to be back on shortly. Ill keep this open

  14. agent0smith
    • 3 years ago
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    Oh and you can find the vertex of a graph y=ax^2+bx+c as \[\Large x=\frac{ -b }{ 2a }\] then you plug in that x value to find y. That will be the minimum value for y (ie the lowest point in the range)

  15. goformit100
    • 3 years ago
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    First of all for solving this question do you know the conception of : Parametric Equations ? And after doing that, I personally guarantee that you can on your own solve this particular question. I personally believe, the happiness of solving Mathematics is much better.

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