Asked by Genny
A model rocket is shot straight up into the air. The table shows its height, h(t), at time t. Graph the table below and determine a function, in standard form, that estimates the height of the rocket at any given time.
Time(s) Height (m)
0 0.0
1 25.1
2 40.4
3 45.9
4 41.6
5 27.5
6 3.6
I graphed it and there is only one x intercept which is (0,0) and the vertex is (3, 45.9) im having trouble determining a function in standard form
Time(s) Height (m)
0 0.0
1 25.1
2 40.4
3 45.9
4 41.6
5 27.5
6 3.6
I graphed it and there is only one x intercept which is (0,0) and the vertex is (3, 45.9) im having trouble determining a function in standard form
Answers
Answered by
Steve
clearly there is an x-intercept somewhere just beyond 6. See how the height is descending.
Since the given h values do not show an obvious symmetry, it is not immediately clear where the vertex is.
However, since h(0) = 0, if
h(t) = at^2 + bt + c
c = 0, so
h(t) = at^2 + bt
Plugging in t=1 and t=2, we have
a+b = 25.1
4a+2b = 40.4
a = -4.9, b=30
h(t) = 30t - 4.9t^2
So, the other t-intercept is at t = 6.12
Since the given h values do not show an obvious symmetry, it is not immediately clear where the vertex is.
However, since h(0) = 0, if
h(t) = at^2 + bt + c
c = 0, so
h(t) = at^2 + bt
Plugging in t=1 and t=2, we have
a+b = 25.1
4a+2b = 40.4
a = -4.9, b=30
h(t) = 30t - 4.9t^2
So, the other t-intercept is at t = 6.12
Answered by
Genny
So what is the vertex?
Answered by
Steve
Huh? You said above what the vertex is.
as always, it is at t = -b/2a
It is not at t=3, but very close. Probably good enough.
as always, it is at t = -b/2a
It is not at t=3, but very close. Probably good enough.
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