anonymous
  • anonymous
@timo86m Is this B? http://prntscr.com/12c0cm
Precalculus
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
http://prntscr.com/12c0hj - thats D
anonymous
  • anonymous
I am looking at it :)
anonymous
  • anonymous
Take your time :)

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agent0smith
  • agent0smith
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)
agent0smith
  • agent0smith
First one isn't B.
anonymous
  • anonymous
Opps! I ment D not B
anonymous
  • anonymous
Let me double check to make sure D is my final answere lol
anonymous
  • anonymous
http://graphsketch.com/parametric.php :D ooh
agent0smith
  • agent0smith
y-intercept is when x=0
anonymous
  • anonymous
well. I know its C now but y-intercept when x=0? what do you mean?
agent0smith
  • agent0smith
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.
agent0smith
  • agent0smith
This would be easier if the graphs weren't so tiny...
anonymous
  • anonymous
I can send them bigger lol. I have to run for a little. I hope to be back on shortly. Ill keep this open
agent0smith
  • agent0smith
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)
goformit100
  • goformit100
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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