Question

solving quadratic equations using common factors practice

during a water balloon fight, Louisa uses a slingshot to fire a water balloon over the fortified wall of the opponent. when fired the slingshot, the balloon will follow the path h(x)=-x=12x

Answer 1

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:

1. \( x = 0 \)
2. \( -x + 12 = 0 \) → \( x = 12 \)

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!