It looks like there's a small error in the function you provided for the height of the water balloon. The correct quadratic function should be in the standard form \( h(x) = ax^2 + bx + c \). Based on what you wrote, you might have meant \( h(x) = -x^2 + 12x \).
Let's assume the function is actually:
\[ h(x) = -x^2 + 12x \]
Now, if we need to solve for when the height \( h(x) = 0 \) (which represents when the water balloon hits the ground), we can set up the equation:
\[ -x^2 + 12x = 0 \]
We can factor out the common term \( x \):
\[ x(-x + 12) = 0 \]
This gives us two solutions:
- \( x = 0 \)
- \( -x + 12 = 0 \) → \( x = 12 \)
Thus, the water balloon hits the ground at \( x = 0 \) and \( x = 12 \).
To summarize, the two points are:
- When the balloon is fired (\( x = 0 \)).
- When the balloon hits the ground (\( x = 12 \)).
Please let me know if you need further assistance or examples!
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