Question 129959: Tutor:
Could you PLEASE help me understand the following problem? My algebra 2 teacher wrote the answer on the board and I copied it down in hopes that I could figure it out later, but apparently I can't.
Icann Flie loves to hanglide at a nearby beach. When he jumps off the escarpment he dives downward toward the water until the wind catches his hanglider, then he flies upward. Assume that his distance in feet from the water varies quadratically with the time since he started his flight. The escarpment is 105 feet above the water. Five seconds after he jumps he is 30 feet above the water. Five seconds later he is 21 feet above the water.
a. Define the variables, write the ordered pairs, and find the particular equation of this function expressing distance above the water in terms of seconds in flight.
The teacher put this on the board (sorry if it is out of order...I don't particularly know where everything goes):
x=time y=distance above water y=ab^x
(0,105) (5,30) (10,21)
y=xt^2+xt+105
answer: y=1.32t^2-21.6t+105
I understand where the teacher got the plots, but after that, I don't know what I'm doing. Log doesn't work the way I'm doing it, and I have no clue what to do with it.
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! quadratically means an equation of the form y=ax^2+bx+c
using the 3 data points, you can generate 3 equations and solve for a, b, and c
the points are (x,y) __ (0,105); (5,30); (10,21)
105=a(0)^2+b(0)+c __ 105=0+0+c __ 105=c __ this was a gift!
30=a(5)^2+b(5)+105 __ -75=25a+5b __ -15=5a+b __1st
21=a(10)^2+b(10)+105 __ -84=100a+10b __ 2nd
multiplying 1st equation by 10 __ -150=50a+10b
subtracting from 2nd equation __ 66=50a __ 1.32=a
substituting __ -15=5(1.32)+b __ -15=6.6+b __ -21.6=b
y=1.32x^2-21.6x+105
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