HELPPP!!!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

Answer:[tex]y=-\frac{1}{3}x^2+2x+5[/tex]Step-by-step explanation:First we are going to get all of these points into our standard form equation for a parabola, which is:[tex]y=ax^2+bx+c[/tex].  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:[tex]8=a(3)^2+b(3)+c[/tex] or [tex]8=9a+3b+c[/tex]Next coordinate pair is x = 5 and y = 20/3, so filling in:[tex]\frac{20}{3}=a(5)^2+b(5)+c[/tex] or [tex]\frac{20}{3}=25a+5b+c[/tex]Next coodinate pair is x = 6 and y = 5, so filling in:[tex]5=a(6)^2+b(6)+c[/tex] or [tex]5=36a+6b+c[/tex]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  =  -5We use the addition/elimination method to eliminate the c's:- 36a  -  6b  -  c  =  -5    9a  +  3b  + c  =  8gives us, when we add:-27a - 3b = 3Now 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 - cWe use the addition/elimination method to eliminate the c's:-9a  -  3b  -  c  =  -825a +  5b  + c  = 20/3gives us, when we add:[tex]16a+2b=-\frac{4}{3}[/tex]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  = -4gives us, when we add:-6a = 2 so [tex]a=-\frac{1}{3}[/tex]Going back to the first bold equation, we will fill in our a value to solve for b:[tex]-27(-\frac{1}{3})-3b=3[/tex] and9 - 3b = 3 and-3b = -6 sob = 2Now 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:[tex]9(-\frac{1}{3})+3(2)+c=8[/tex] and-3 + 6 + c = 8 and3 + c = 8 soc = 5Our equation, then is:[tex]y=-\frac{1}{3}x^2+2x+5[/tex]