a y=-16x^2+33x+46, where x is the number of seconds after throwing the ball and y is the height.
B
b 28.5ft
C
c About 63ft.
Practice makes perfect
a To find the equation of the quadratic function, we will use the values in the table.
Time (s)
Height (ft)
0
46
1
63
2
48
3
1
We will substitute the first three points onto the standard form of a quadratic equation, namely, y=ax^2+bx+c.
x
y
Equation
0
46
46=a(0)^2+b(0)+c
1
63
63=a(1)^2+b(1)+c
2
48
48=a(2)^2+b(2)+c
After simplifying and rearranging the equations, we get the system below.
c=46 & (I) a+b+c=63 & (II) 4a+2b+c=48 & (III)
From the first equation we already got the value of c. So, we actually have a system of two equations. Let's substitute c=46 into the last two equations and simplify them.
a+b+46=63 4a+2b+46=48
⇒
a+b=17 2a+b=1
Next, we subtract the second equation from the first one.
a+b &= 17
^- 2a+b &= 1
- a &= 16
From the latter equation, we get a=-16. Let's substitute it into the first equation of the final system to find the value of b.
-16 +b = 17 ⇒ b = 33
Consequently, the quadratic equation that models the given data is y=-16x^2+33x+46.
b To know the height of the ball at 2.5 seconds we have to evaluate the function we obtained in Part A at x=2.5.
Therefore, at 2.5 seconds the ball was at a height of 28.5ft.
c The function that models the path of the ball is y=-16x^2+33x+46. Since a=-16< 0, the parabola opens downward. This implies that the vertex corresponds to the maximum height of the ball. Let's find the x-coordinate of the vertex.
x=-b/2a ⇒ x = -33/2(-16) = 1.031 25
To find the y-coordinate of the vertex, we substitute x=1.031 25 into the quadratic function.
