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Skittles456
 one year ago
Best ResponseYou've already chosen the best response.0http://prntscr.com/12c0hj  thats D

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1For 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
 one year ago
Best ResponseYou've already chosen the best response.1First one isn't B.

Skittles456
 one year ago
Best ResponseYou've already chosen the best response.0Opps! I ment D not B

Skittles456
 one year ago
Best ResponseYou've already chosen the best response.0Let me double check to make sure D is my final answere lol

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1yintercept is when x=0

Skittles456
 one year ago
Best ResponseYou've already chosen the best response.0well. I know its C now but yintercept when x=0? what do you mean?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1The yintercept is... when x=0. Did you find the yintercept 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 yintercept, so you'll have to plug in a couple of points to check which is correct.

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1This would be easier if the graphs weren't so tiny...

Skittles456
 one year ago
Best ResponseYou've already chosen the best response.0I can send them bigger lol. I have to run for a little. I hope to be back on shortly. Ill keep this open

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0First 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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