SOLUTION: You throw a ball straight up from a rooftop. The ball misses the rooftop on its way down and eventually strikes the ground. A mathematical model can be used to describe the relatio

Algebra ->  Linear-equations -> SOLUTION: You throw a ball straight up from a rooftop. The ball misses the rooftop on its way down and eventually strikes the ground. A mathematical model can be used to describe the relatio      Log On


   



Question 1124230: You throw a ball straight up from a rooftop. The ball misses the rooftop on its way down and eventually strikes the ground. A mathematical model can be used to describe the relationship for the ball's height above the ground, y, after x seconds.
Below is the table:
x, seconds after the ball is thrown. y, ball's height, infeet, above the ground
1 339
3 321
4 264
Find the quadratic function y= ax2 + bx + c whose graph passes through the given points.
Y=

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Below is the table:
x- {seconds after the ball is thrown)
+y-(ball's height, infeet, above the ground)


x |y
1%09 339
3 321
4%09 +264
Find the quadratic function y=+ax%5E2+%2B+bx+%2B+c whose graph passes through the given points.
y=+ax%5E2+%2B+bx+%2B+c...plug in x=1,y=339
339=+a%2A1%5E2+%2B+b%2A1+%2B+c
339=+a+%2B+b+%2B+c........eq.1

y=+ax%5E2+%2B+bx+%2B+c...plug in x=3,y=321
321=+a%2A3%5E2+%2B+b%2A3+%2B+c
321=+9a+%2B+3b+%2B+c........eq.2

y=+ax%5E2+%2B+bx+%2B+c...plug in x=4,y=264
264=+a%2A4%5E2+%2B+b%2A4+%2B+c
264=+16a+%2B+4b+%2B+c........eq.3
------------------------------solve the system
339=+a+%2B+b+%2B+c........eq.1
321=+9a+%2B+3b+%2B+c........eq.2
-----------------------------------------subtract eq2 from eq1
339-321=+a-9a+%2B+b-3b+%2B+c-c
18=+-8a+-2b+....solve for a
8a=++-2b-18+
a=++-2b%2F8-18%2F8+
a=++-b%2F4-9%2F4+...............eq.1a
go to
264=+16a+%2B+4b+%2B+c........eq.3
321=+9a+%2B+3b+%2B+c........eq.2
-----------------------------------------------subtract eq.2 rom eq.3
264-321=+16a-9a+%2B+4b-3b+%2B+c-c
-57=+7a+%2B+b
-57-b=+7a+
a=-57%2F7-b%2F7+...............eq.1b

from eq.1a and eq.1b we have:
+-b%2F4-9%2F4=+-57%2F7-b%2F7............solve for b
+%28-b-9%29%2F4=+%28-57-b%29%2F7
7%28-b-9%29=+4%28-57-b%29
-7b-63=+-228-4b
228-63=+7b-4b
165=+3b
+b=165%2F3
highlight%28+b=+55%29
go back to a=++-b%2F4-9%2F4+...............eq.1a, substitute b
a=++-%2855%29%2F4-9%2F4+
a=++-%2855%2B9%29%2F4+
a=++-64%2F4+
highlight%28a=++-16%29+

go to 339=+a+%2B+b+%2B+c........eq.1, substitute a and b
339=+-16+%2B55+%2B+c
339=+39%2B+c
c=339-39
highlight%28c=300%29
then, your equation is:
highlight%28y=+-16x%5E2+%2B55x+%2B+300+%29