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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

3 answers

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.

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

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.