Question

1. xenia launches a water balloon into the air. the balloon reaches a maximum height of 40 feet above the ground and is in the air for 2 seconds. create a quadratic equation that models h, the height of the balloon, in feet, x seconds after it is launched from the ground.

Explanation:

Step1: Identify the vertex form of a quadratic

The vertex form of a quadratic equation is \( h(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Here, the maximum height (vertex) occurs at \( x = 2 \) seconds (since it reaches max height in 2 seconds) and \( k = 40 \) feet. So the equation becomes \( h(x) = a(x - 2)^2 + 40 \).

Step2: Determine the value of \( a \)

We know the balloon is launched from the ground, so when \( x = 0 \), \( h(0) = 0 \). Substitute \( x = 0 \) and \( h(0) = 0 \) into the equation:
\( 0 = a(0 - 2)^2 + 40 \)
\( 0 = 4a + 40 \)
Subtract 40 from both sides: \( -40 = 4a \)
Divide by 4: \( a = -10 \)

Step3: Write the final equation

Substitute \( a = -10 \), \( h = 2 \), and \( k = 40 \) into the vertex form:
\( h(x) = -10(x - 2)^2 + 40 \)
We can expand this to standard form:
\( h(x) = -10(x^2 - 4x + 4) + 40 \)
\( h(x) = -10x^2 + 40x - 40 + 40 \)
\( h(x) = -10x^2 + 40x \)

Answer:

The quadratic equation modeling the height is \( \boldsymbol{h(x) = -10x^2 + 40x} \) (or in vertex form \( \boldsymbol{h(x) = -10(x - 2)^2 + 40} \)).