Question

Use the table to answer the question.

x Substituted Evaluate True Statement?
12
24
36
48
When set off, a certain firework follows the path of the quadratic function h=−2536x2+1623x, where:

h= the height of the firework in feet.
x= the horizontal distance it travels in feet.
To determine how far the firework will travel before reaching the ground, determine which value of x in table is a solution to the equation 0=−2536x2+1623x.

(1 point)
Responses

12 feet
12 feet

48 feet
48 feet

24 feet
24 feet

36 feet

Answer 1

To determine how far the firework will travel before reaching the ground, we need to solve the equation \(0 = -2536x^2 + 1623x\). This can be factored as follows:

1. Factor out \(x\):
\[
0 = x(-2536x + 1623)
\]

This gives us two possible solutions:
1. \(x = 0\) (the firework starts at the launch point)
2. \(-2536x + 1623 = 0\)

Now, let's solve for \(x\) in the second equation:
\[
-2536x + 1623 = 0
\]
\[
2536x = 1623
\]
\[
x = \frac{1623}{2536}
\]
Calculating that gives:
\[
x \approx 0.640
\]

Now we will check the proposed values \(12, 24, 36,\) and \(48\) feet to see if they solve the equation \(0 = -2536x^2 + 1623x\).

1. For \(x = 12\):
\[
0 = -2536(12^2) + 1623(12) \\
= -2536(144) + 1623(12) \\
= -364704 + 19476 \\
= -345228 \quad (\text{not } 0)
\]

2. For \(x = 24\):
\[
0 = -2536(24^2) + 1623(24) \\
= -2536(576) + 1623(24) \\
= -1457280 + 38952 \\
= -1418328 \quad (\text{not } 0)
\]

3. For \(x = 36\):
\[
0 = -2536(36^2) + 1623(36) \\
= -2536(1296) + 1623(36) \\
= -3282176 + 58428 \\
= -3227748 \quad (\text{not } 0)
\]

4. For \(x = 48\):
\[
0 = -2536(48^2) + 1623(48) \\
= -2536(2304) + 1623(48) \\
= -5830656 + 77844 \\
= -5752812 \quad (\text{not } 0)
\]

Since all tested values did not satisfy the equation, it appears that the values mentioned in the question do not include any solutions.

However, assuming \(x\) approaches the x-intercept where the firework reaches the ground, it can be inferred that it travels a maximum horizontal distance of just under those values or up to \(x = \frac{1623}{2536}\), which is not listed.

If you're looking for the best guess from the provided choices, \(48\) feet would be the furthest distance before the firework returns to the ground, considering similar quadratic equations commonly yield max height and ground return at higher values.

Therefore, the most reasonable answer given the options could be:
**48 feet**