Question

Path of a soccer ball is modeled by h(x)= -1/2x(x-20) You make the equation 0=-1/2x(x-20) How can the zero-product property help find the horizontal distance the ball has traveled when the h equals to zero.
So I would say -1/2 x=0 and x-20=0
So, this would be x=0 and x=20
The answer would be that I have to solve both equations

Answer 1

Thanks.
The following are the possible answers:
1. solve both equations
2. Plug 0 for all the x in the function.
3. Cancel out x then multiply -1/2 and -20

Answer 2

You're on the right track in using the zero-product property to solve the equation \( h(x) = 0 \) based on the given model of the soccer ball's path:

\[
0 = -\frac{1}{2}x(x - 20)
\]

To apply the zero-product property, we can simplify the equation by ignoring the constant factor of \(-\frac{1}{2}\) (since it doesn't affect the zeros of the equation). Thus, we can rewrite the equation as:

\[
0 = x(x - 20)
\]

Setting this equal to zero gives us two factors to solve for:

1. \( x = 0 \)
2. \( x - 20 = 0 \) ⇒ \( x = 20 \)

These solutions tell us that the horizontal distance \( x \) where the height \( h(x) = 0 \) (i.e., when the soccer ball hits the ground) occurs at:

- \( x = 0 \): This represents the point when the ball was initially kicked (at the origin).
- \( x = 20 \): This indicates the point where the ball lands (20 units away horizontally).

In conclusion, the horizontal distances where the ball is on the ground (height \( h = 0 \)) are when \( x = 0 \) and \( x = 20 \). Thus, the ball has traveled a distance of 20 units before it lands.

Answer 3

To find the horizontal distance the soccer ball has traveled when \( h(x) = 0 \) using the equation \( 0 = -\frac{1}{2}x(x - 20) \), the best approach is to **solve both equations** derived from the zero-product property.

### Let's evaluate the options:

1. **Solve both equations**: This is the correct choice. By setting \( 0 = x(x - 20) \), you can directly use the zero-product property to find the solutions \( x = 0 \) and \( x = 20 \). This tells you the initial position (when the ball is kicked) and where it lands.

2. **Plug 0 for all the x in the function**: While this will help you evaluate the function at \( x = 0 \) (and you'll see that \( h(0) = 0 \)), it doesn’t address the full problem of finding the other point where the ball hits the ground. This method is incomplete for finding all \( x \) where \( h(x) = 0 \).

3. **Cancel out x then multiply -1/2 and -20**: This option suggests a misunderstanding of the zero-product property. You should not cancel \( x \) without understanding the implications, as doing so when \( x = 0 \) can lead to a loss of one of the solutions. Specifically, canceling could give the impression that \( x \) cannot be zero, which is incorrect since that represents the starting point.

### Conclusion:

The best option to find the points where the ball hits the ground is:
**1. Solve both equations.**

This approach accurately provides both values of \( x \) (0 and 20) where the height \( h(x) \) is zero.