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Skittles456Best ResponseYou've already chosen the best response.0
http://prntscr.com/12c0hj  thats D
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
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)
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
First one isn't B.
 one year ago

Skittles456Best ResponseYou've already chosen the best response.0
Opps! I ment D not B
 one year ago

Skittles456Best ResponseYou've already chosen the best response.0
Let me double check to make sure D is my final answere lol
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
http://graphsketch.com/parametric.php :D ooh
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
yintercept is when x=0
 one year ago

Skittles456Best ResponseYou've already chosen the best response.0
well. I know its C now but yintercept when x=0? what do you mean?
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
The 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.
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
This would be easier if the graphs weren't so tiny...
 one year ago

Skittles456Best ResponseYou've already chosen the best response.0
I can send them bigger lol. I have to run for a little. I hope to be back on shortly. Ill keep this open
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
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)
 one year ago

goformit100Best ResponseYou've already chosen the best response.0
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.
 one year ago
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