Question

A group of students launches a model rocket from the top of a building the students have determined the equation y= -1/2(x + 6) (x -18) to describe the height of the rocket above ground level y in metres at any ti
W x in seconds.
table with 0,2,4,6,8,10,12,14,16,18 what are the heights above ground y.

Use the equation ti determine zeros
Determine the equation if the axis of symmetry
Determine the maximum height if the rocket

Answer 1

To analyze the model rocket's height given by the equation \( y = -\frac{1}{2}(x + 6)(x - 18) \), let’s tackle each part step-by-step.

### 1. Height Calculation at Given Time Intervals
We'll calculate the height \( y \) for each time \( x \) in the table: 0, 2, 4, 6, 8, 10, 12, 14, 16, 18.

**Equation:**
\[ y = -\frac{1}{2}(x + 6)(x - 18) \]

**Calculations:**

- \( x = 0 \):
\[ y = -\frac{1}{2}(0 + 6)(0 - 18) = -\frac{1}{2}(6)(-18) = 54 \]

- \( x = 2 \):
\[ y = -\frac{1}{2}(2 + 6)(2 - 18) = -\frac{1}{2}(8)(-16) = 64 \]

- \( x = 4 \):
\[ y = -\frac{1}{2}(4 + 6)(4 - 18) = -\frac{1}{2}(10)(-14) = 70 \]

- \( x = 6 \):
\[ y = -\frac{1}{2}(6 + 6)(6 - 18) = -\frac{1}{2}(12)(-12) = 72 \]

- \( x = 8 \):
\[ y = -\frac{1}{2}(8 + 6)(8 - 18) = -\frac{1}{2}(14)(-10) = 70 \]

- \( x = 10 \):
\[ y = -\frac{1}{2}(10 + 6)(10 - 18) = -\frac{1}{2}(16)(-8) = 64 \]

- \( x = 12 \):
\[ y = -\frac{1}{2}(12 + 6)(12 - 18) = -\frac{1}{2}(18)(-6) = 54 \]

- \( x = 14 \):
\[ y = -\frac{1}{2}(14 + 6)(14 - 18) = -\frac{1}{2}(20)(-4) = 40 \]

- \( x = 16 \):
\[ y = -\frac{1}{2}(16 + 6)(16 - 18) = -\frac{1}{2}(22)(-2) = 22 \]

- \( x = 18 \):
\[ y = -\frac{1}{2}(18 + 6)(18 - 18) = -\frac{1}{2}(24)(0) = 0 \]

### Summary of Heights
| Time \(x\) | Height \(y\) |
|------------|--------------|
| 0 | 54 |
| 2 | 64 |
| 4 | 70 |
| 6 | 72 |
| 8 | 70 |
| 10 | 64 |
| 12 | 54 |
| 14 | 40 |
| 16 | 22 |
| 18 | 0 |

### 2. Determine Zeros
Zeros of the equation occur when \( y = 0 \):
\[ -\frac{1}{2}(x + 6)(x - 18) = 0 \]

Setting each factor equal to zero:
1. \( x + 6 = 0 \) → \( x = -6 \)
2. \( x - 18 = 0 \) → \( x = 18 \)

**The zeros are at \( x = -6 \) and \( x = 18 \)**.

### 3. Determine the Axis of Symmetry
The general form for the vertex form of a quadratic function \( ax^2 + bx + c \) has an axis of symmetry given by:
\[ x = -\frac{b}{2a} \]

To convert:
\[ y = -\frac{1}{2}(x^2 - 12x - 108) \]
This expands to \( y = -\frac{1}{2}x^2 + 6x + 54 \).

Here, \( a = -\frac{1}{2} \) and \( b = 6 \):

\[ x = -\frac{6}{2 \cdot -\frac{1}{2}} = -\frac{6}{-1} = 6 \]

**The axis of symmetry is \( x = 6 \)**.

### 4. Determine Maximum Height
To find the maximum height, substitute the axis of symmetry value \( x = 6 \) into the height formula:

\[ y(6) = -\frac{1}{2}(6 + 6)(6 - 18) \]
\[ y(6) = -\frac{1}{2}(12)(-12) = 72 \]

**The maximum height of the rocket is \( 72 \) meters**.