imagine that as a ball is tossed, its motion is tracked on a coordinate plane. given only a few of the points that the ball passes through, it is actually possible to determine the equation of the parabola that represents the ball’s path through the air.
answer part a in the picture

Step-by-step explanation:

First we are going to get all of these points into our standard form equation for a parabola, which is:

.  We will do this by using the x and y coordinates from each of those 3 coordinate pairs.  In the first pair, x = 3 and y = 8, so filling in:

or

Next coordinate pair is x = 5 and y = 20/3, so filling in:

or

Next coodinate pair is x = 6 and y = 5, so filling in:

or

We will use the first and the third equations and multiply one of them through by a -1 to get rid of the c's:

-1(36a  +  6b  +  c  =  5) = -35a  -  6b  -  c  =  -5

We use the addition/elimination method to eliminate the c's:

- 36a  -  6b  -  c  =  -5

   9a  +  3b  + c  =  8

gives us, when we add:

-27a - 3b = 3

Now we will do the same thing with the first and the second equations to eliminate the c's.  We multiply one of them through by a -1:

-1(8 = 9a + 3b + c) = -8 = -9a - 3b - c

We use the addition/elimination method to eliminate the c's:

-9a  -  3b  -  c  =  -8

25a +  5b  + c  = 20/3

gives us, when we add:

Now we will use the addition/elimination method on those 2 bold equations, multiplying through by what we need, in order to solve for a.  The first bold equation will be multiplied through by 2, and the second bold equation will be multiplied through by 3 to give us 2 new equations that are:

-54a  -  6b  =  6

48a  +  6b  = -4

gives us, when we add:

-6a = 2 so

Going back to the first bold equation, we will fill in our a value to solve for b:

and

9 - 3b = 3 and

-3b = -6 so

b = 2

Now we will go back up to the very first equation we wrote in terms of a, b, and c and fill in a and b to solve for c:

and

-3 + 6 + c = 8 and

3 + c = 8 so

c = 5

Our equation, then is: