Question

terry throws a ball into a hoop that is 5 feet off the ground. the path of the ball can be modeled by the equation $y = -x^2 + 4x + 5$, where $x$ represents the horizontal distance traveled by the ball, in feet, and $y$ represents the vertical height of the ball, in feet. since the ball hoop is located 5 feet off the ground, you can use the equation $y = 5$ to model the position of the hoop. determine the horizontal distance between terry and the hoop. (1 point)

\bigcirc 2 feet

\bigcirc 0 feet

\bigcirc 4 feet

\bigcirc 5 feet

Explanation:

Step1: Set y = 5 in the equation

We know the ball's path is \( y = -x^2 + 4x + 5 \) and the hoop is at \( y = 5 \). So substitute \( y = 5 \) into the equation:
\( 5 = -x^2 + 4x + 5 \)

Step2: Solve the quadratic equation

Subtract 5 from both sides:
\( 0 = -x^2 + 4x \)
Factor out \( -x \):
\( 0 = -x(x - 4) \)

This gives two solutions: \( -x = 0 \) (so \( x = 0 \)) or \( x - 4 = 0 \) (so \( x = 4 \)).

\( x = 0 \) is Terry's starting position (horizontal distance 0 from himself). The hoop is at \( x = 4 \), so the horizontal distance between Terry (at \( x = 0 \)) and the hoop (at \( x = 4 \)) is \( 4 - 0 = 4 \) feet? Wait, no—wait, let's check again. Wait, when \( y = 5 \), solving \( 5 = -x^2 + 4x + 5 \) simplifies to \( 0 = -x^2 + 4x \), so \( x^2 - 4x = 0 \), \( x(x - 4) = 0 \), so \( x = 0 \) or \( x = 4 \). But Terry is at \( x = 0 \) (throwing the ball), so the hoop is at \( x = 4 \)? Wait, but the options include 4 feet. Wait, but let's re-express the equation. Wait, maybe I made a mistake. Wait, the equation is \( y = -x^2 + 4x + 5 \). When \( y = 5 \), \( 5 = -x^2 + 4x + 5 \) → \( 0 = -x^2 + 4x \) → \( x^2 - 4x = 0 \) → \( x(x - 4) = 0 \), so \( x = 0 \) or \( x = 4 \). So the horizontal distance from Terry (x=0) to the hoop (x=4) is 4 - 0 = 4 feet? But wait, the options have 4 feet as an option. Wait, but let's check the options again. The options are 2, 0, 4, 5. Wait, maybe I misread the equation. Wait, the equation is \( y = -x^2 + 4x + 5 \). Let's complete the square: \( y = -(x^2 - 4x) + 5 = -(x^2 - 4x + 4 - 4) + 5 = -(x - 2)^2 + 4 + 5 = -(x - 2)^2 + 9 \). So the vertex is at (2, 9). But when \( y = 5 \), solving \( 5 = -(x - 2)^2 + 9 \) → \( (x - 2)^2 = 4 \) → \( x - 2 = \pm 2 \) → \( x = 2 + 2 = 4 \) or \( x = 2 - 2 = 0 \). So the two points where y=5 are x=0 and x=4. So Terry is at x=0 (throwing the ball), so the hoop is at x=4, so the horizontal distance is 4 feet. Wait, but the option is 4 feet. So the answer should be 4 feet? Wait, but let me check again. Wait, maybe I messed up the substitution. Wait, the problem says "the horizontal distance between Terry and the hoop". Terry is at x=0 (since when x=0, that's his position, as he throws the ball from there). The hoop is at x=4, so distance is 4 - 0 = 4 feet. So the correct option is 4 feet. Wait, but let's check the options: the options are 2, 0, 4, 5. So 4 feet is an option. So the answer is 4 feet. Wait, but wait—maybe I made a mistake. Wait, let's solve \( 5 = -x^2 + 4x + 5 \):

Subtract 5: \( 0 = -x^2 + 4x \)

Multiply both sides by -1: \( 0 = x^2 - 4x \)

Factor: \( x(x - 4) = 0 \)

Solutions: x=0 or x=4. So the hoop is at x=4, Terry at x=0, so distance is 4 - 0 = 4 feet. So the answer is 4 feet. Wait, but the option is "4 feet". So the correct option is 4 feet. Wait, but let me check again. Wait, maybe the equation is different? Wait, the original equation is \( y = -x^2 + 4x + 5 \). Yes. So when y=5, x=0 or x=4. So the horizontal distance is 4 feet. So the answer is 4 feet. Wait, but the options include 4 feet. So the correct option is 4 feet. Wait, but let me check the options again. The options are:

- 2 feet

- 0 feet

- 4 feet

- 5 feet

So the correct answer is 4 feet. Wait, but wait—maybe I made a mistake. Wait, when x=0, y=5? Wait, when x=0, y = -0 + 0 + 5 = 5. Oh! Wait a minute! Terry throws the ball into a hoop that is 5 feet off the ground. If at x=0, y=5, that means Terry is at the hoop? But that can't be. Wait, this is a mistake. Wait, when x=0, y = -0 + 0 + 5 = 5. So Terry is at (0,5), and the hoop is also at y=5. So solving for y=5 gives x=0 and x=4? Wait, no—wait, when x=0, y=5. So Terry is at (0,5), and the hoop is at (4,5)? But that would mean the hoop is 4 feet horizontally from Terry. But if Terry is at (0,5), and the hoop is at (4,5), then the horizontal distance is 4 feet. But wait, when x=0, y=5, so Terry is at 5 feet height? But the hoop is 5 feet off the ground. So maybe Terry is at ground level? Wait, maybe the equation's y is the height above ground, and x is horizontal distance from Terry. So when x=0, that's Terry's position (horizontal distance 0), and y=5 is his height? No, that doesn't make sense. Wait, maybe the equation is modeled with x=0 being Terry's position, and y is the height. So when he throws the ball, at x=0, y=5 (his hand is 5 feet off the ground?), and the hoop is also 5 feet off the ground. So we need to find the x where y=5, other than x=0 (since x=0 is Terry's position). So solving \( 5 = -x^2 + 4x + 5 \) gives x=0 and x=4. So the hoop is at x=4, so the horizontal distance is 4 feet. So the answer is 4 feet.

Answer:

4 feet (corresponding to the option "4 feet")