SOLUTION: I have been asked to draw a position tme graph based on the following data which details a fairground ride in the first 8 seconds - but my mind has gone completely blank. 1 y =

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: I have been asked to draw a position tme graph based on the following data which details a fairground ride in the first 8 seconds - but my mind has gone completely blank. 1 y =       Log On

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Question 4893: I have been asked to draw a position tme graph based on the following data which details a fairground ride in the first 8 seconds - but my mind has gone completely blank.
1 y = 4t^2 for first 2 seconds
2 y = -4t^2 + 32t - 32 from 2 - 6 seconds
3 y = 4t^2 - 64t + 256 from 6 - 8 seconds
Any help appreciated.

Answer by Abbey(339) About Me  (Show Source):
You can put this solution on YOUR website!
Any squared term is a parabola... In this case it begins at zero, because you don't have a negative time
Treat your x axis as the t in this case, plug in a couple of numbers for the first equation:
If t=0, y = 0 (Plot (0,0))
If t=1, y = 4 (Plot (1,4))
Connect these two with a curving line (like the bottom of a bowl)
the second equation begins at 2 seconds and will be shaped like an upside down bowl
-4t^2+32t-32
If t=2, then y=-48 (plot 2,-48)
-4(2)^2+32(2)-32=-48
If t=6 then y=16 (plot 6,16)
-4(6)^2+32(6)-32=16
The third one will have the same shape as the first: a parabola opening upward, extending to the right:
4(t)^2-64t+256
If t=6 then y=16 (plot(6,16))
If t=8 then y=0 (plot (8,0))
You can always just plug more numbers into your equation if you are unsure of the shape of it.