
1 point
Find out the equation of the curve followed by the firework.

y
−
15
=
−
1
15
x
2
y−15=− 
15
1
​
 x 
2
 

y
−
15
=
−
1
15
(
x
−
15
)
2
y−15=− 
15
1
​
 (x−15) 
2
 

y
−
30
=
−
1
30
(
x
−
30
)
2
y−30=− 
30
1
​
 (x−30) 
2
 

y
−
30
=
−
1
30
x
2
y−30=− 
30
1
​
 x 
2

To find the correct equation of the curve followed by the firework, let's analyze the given options:

### Key Details:
- The firework is launched from the origin \( (0, 0) \) and reaches a maximum height of 15 meters at \( x = 15 \) meters.
- The parabola representing the firework’s path opens downward, since it reaches a maximum height.

Given these points, the vertex form of a parabola is:

\[
y = a(x - h)^2 + k
\]

where \( (h, k) \) is the vertex. For the firework:
- \( h = 15 \)
- \( k = 15 \)

So, the vertex form becomes:

\[
y = a(x - 15)^2 + 15
\]

Since the firework starts at the origin \( (0, 0) \) and follows the path back to the ground after reaching the maximum height, the correct vertex form (using \( y = 0 \) when \( x = 0 \)) becomes:

\[
y - 15 = a(x - 15)^2
\]

The value of \( a \) can be determined by using the condition that when \( x = 0 \), \( y = 0 \):

\[
0 - 15 = a(0 - 15)^2
\]

\[
-15 = a(225)
\]

\[
a = -\frac{1}{15}
\]

So the correct equation is:

\[
y - 15 = -\frac{1}{15}(x - 15)^2
\]

### Evaluating the Options:
1. \( y - 15 = -\frac{1}{15}x^2 \)
2. \( y - 15 = -\frac{1}{15}(x - 15)^2 \) **(Correct)**
3. \( y - 30 = -\frac{1}{30}(x - 30)^2 \)
4. \( y - 30 = -\frac{1}{30}x^2 \)

### Correct Answer:
The correct equation of the curve followed by the firework is:

\[
y - 15 = -\frac{1}{15}(x - 15)^2
\]

Let me know if you need more details or have any other questions!

### Relative Questions:
1. How would the equation change if the firework had a different maximum height?
2. What effect does changing the horizontal distance of the landing point have on the equation?
3. Can you derive the equation of the firework's path if it lands 40 meters away instead?
4. How does the coefficient of \( x^2 \) determine the width of the parabola?
5. How would the equation differ if the vertex were at a different height, say 20 meters?

**Tip:** For any projectile motion described by a parabola, the vertex form is a powerful tool to quickly derive the equation based on the maximum height and the symmetry of the path.

Learn how to find the equation of the path followed by a firework using vertex form of a parabola. This problem involves determining the vertex and coefficient of a parabola based on its maximum height. Suitable for high school students.

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