We are given the function representing the height of an object launched from a building:

$f(x) = -16x^2 + 16x + 96$

where $f(x)$ represents the height in feet, and $x$ is the time in seconds after the launch.

1. Find and Interpret $f(-2)$:

Substitute $x = -2$ into the equation:

$f(-2) = -16(-2)^2 + 16(-2) + 96$

$f(-2) = -16(4) - 32 + 96$ $f(-2) = -64 - 32 + 96 = 0$

Interpretation:
$f(-2) = 0$, meaning that 2 seconds before the object was launched, it was 0 feet above the ground (the object had not yet been launched). In this context, negative time doesn't make sense physically because time starts at the moment of launch, so this is outside of the domain for real-world interpretation.

2. Find and Interpret $f(0.5)$:

Substitute $x = 0.5$ into the equation:

$f(0.5) = -16(0.5)^2 + 16(0.5) + 96$ $f(0.5) = -16(0.25) + 8 + 96$ $f(0.5) = -4 + 8 + 96 = 100$

Interpretation:
$f(0.5) = 100$, meaning that 0.5 seconds after the object was launched, it was 100 feet above the ground. This is a valid, real-world interpretation as it represents the object's height shortly after launch.

3. Find and Interpret $f(6)$:

Substitute $x = 6$ into the equation:

$f(6) = -16(6)^2 + 16(6) + 96$ $f(6) = -16(36) + 96 + 96$ $f(6) = -576 + 96 + 96 = -384$

Interpretation:
$f(6) = -384$, meaning that 6 seconds after the object was launched, it was 384 feet below the ground. Since the height cannot be negative in this context (the object hit the ground and fell), this is outside the valid range for the height function. Therefore, this is not a meaningful value in the real-world scenario.

Appropriate Domain for the Function:

Based on the observations above, the object was launched at $x = 0$, and we need to focus on times when the object is above the ground. The valid domain would be the time interval during which the object is in the air, which ends when $f(x) = 0$ (i.e., when the object hits the ground).

To find when the object hits the ground, solve for $x$ when $f(x) = 0$:

$-16x^2 + 16x + 96 = 0$

We can divide through by -16 to simplify:

$x^2 - x - 6 = 0$

Factor the quadratic:

$(x - 3)(x + 2) = 0$

So, $x = 3$ or $x = -2$.

Since $x = -2$ is a negative time, we discard it, leaving $x = 3$ as the time when the object hits the ground.

Thus, the appropriate domain is:

$0 \leq x \leq 3$

This means the function is valid from the time the object is launched (at $x = 0$) until it hits the ground (at $x = 3$).

Summary:

• $f(-2) = 0$, meaning the object was at ground level before launch, which is outside the valid domain.
• $f(0.5) = 100$, meaning 0.5 seconds after launch, the object was 100 feet above the ground.
• $f(6) = -384$, meaning 6 seconds after launch, the object was below the ground level (not physically valid).
• The appropriate domain is $0 \leq x \leq 3$.

Would you like more details or have any questions on the domain or function interpretation?

Follow-up Questions:

1. How do you find the vertex of a quadratic function, and how does it relate to the object's maximum height?
2. How would you calculate the object's velocity at a specific time?
3. What would the function look like if the object was launched upward at a different initial velocity?
4. How can we graph this quadratic function to visualize the object's flight path?
5. How would the domain change if the object was launched from a higher building?

Tip: Remember that the physical domain of motion problems (like projectile motion) often restricts the time to when the object is above the ground, so always check for when the height reaches zero!