The problem asks us to graph the quadratic equation $y = -x^2 + 10x - 9$. We are asked to make a table of values, carefully following the order of operations to calculate the $y$ values for different $x$ values. Also, interpret the graph as the trajectory of a water balloon launched from a catapult at the one-yard line on a football field, and determine how high the balloon went and where it landed.

1. Problem Description

The problem asks us to graph the quadratic equation $y = -x^2 + 10x - 9$. We are asked to make a table of values, carefully following the order of operations to calculate the $y$ values for different $x$ values. Also, interpret the graph as the trajectory of a water balloon launched from a catapult at the one-yard line on a football field, and determine how high the balloon went and where it landed.

2. Solution Steps

First, we need to create a table of $x$ and $y$ values for the given equation $y = -x^2 + 10x - 9$. We can choose various $x$ values to calculate corresponding $y$ values. To get a good picture of the parabola, we should consider values around the vertex. The $x$-coordinate of the vertex is given by $x = \frac{-b}{2a}$, where $a = -1$ and $b = 10$ in the given equation. So, $x = \frac{-10}{2(-1)} = 5$. Thus, we should choose $x$ values around

5.

Let's choose the following $x$ values: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,

1

0. Now, let's calculate the $y$ values for each $x$ value:

- $x=0$: $y = -(0)^2 + 10(0) - 9 = -9$

- $x=1$: $y = -(1)^2 + 10(1) - 9 = -1 + 10 - 9 = 0$

- $x=2$: $y = -(2)^2 + 10(2) - 9 = -4 + 20 - 9 = 7$

- $x=3$: $y = -(3)^2 + 10(3) - 9 = -9 + 30 - 9 = 12$

- $x=4$: $y = -(4)^2 + 10(4) - 9 = -16 + 40 - 9 = 15$

- $x=5$: $y = -(5)^2 + 10(5) - 9 = -25 + 50 - 9 = 16$

- $x=6$: $y = -(6)^2 + 10(6) - 9 = -36 + 60 - 9 = 15$

- $x=7$: $y = -(7)^2 + 10(7) - 9 = -49 + 70 - 9 = 12$

- $x=8$: $y = -(8)^2 + 10(8) - 9 = -64 + 80 - 9 = 7$

- $x=9$: $y = -(9)^2 + 10(9) - 9 = -81 + 90 - 9 = 0$

- $x=10$: $y = -(10)^2 + 10(10) - 9 = -100 + 100 - 9 = -9$

Now, we have the following table:

x | y

--|--

0 | -9

1 | 0

2 | 7

3 | 12

4 | 15

5 | 16

6 | 15

7 | 12

8 | 7

9 | 0

10| -9

The graph of the water balloon's trajectory is a parabola opening downward. The highest point the balloon reached is the vertex of the parabola, which has coordinates (5, 16). So, the maximum height is 16 yards.

The water balloon lands when the height $y$ is

0. From the table, we see that this happens when $x=1$ and $x=9$. Since the balloon was launched from the one-yard line, x represents the distance from that line. Therefore, the balloon landed at $x=9$ yards from the one-yard line. Since the balloon was launched from the one-yard line, it travelled $9 - 1 = 8$ yards from its launch point. Therefore, the landing spot is 9 yards from the zero yard line.

The vertex of the parabola represents the maximum height, and the $x$-intercepts represent where the balloon lands (height = 0).

3. Final Answer

The water balloon went to a maximum height of 16 yards. The water balloon landed 9 yards from the zero yard line on the football field (or 8 yards from the point it was launched). The vertex of the parabola tells us the maximum height, and the $x$-intercepts tell us where the balloon lands.