Question

1. adrian bounces a tennis ball hard off the ground. the ball is in the air for 6 seconds and reaches a maximum height of 12 feet above the ground. create a quadratic equation that models h, the height of the ball, in feet, x seconds after it bounces off the ground.

Explanation:

Step1: Identify the vertex form of quadratic

The vertex form of a quadratic function is \( h(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. The maximum height is 12 feet, so \( k = 12 \). The ball is in the air for 6 seconds, so it starts at \( x = 0 \) (height 0) and ends at \( x = 6 \) (height 0). The vertex (maximum point) is at the midpoint of \( x = 0 \) and \( x = 6 \), so \( h = \frac{0 + 6}{2} = 3 \). Thus, the vertex is \((3, 12)\), so the equation becomes \( h(x) = a(x - 3)^2 + 12 \).

Step2: Find the value of \( a \)

We know the ball is at height 0 when \( x = 0 \). Substitute \( x = 0 \) and \( h(0) = 0 \) into the equation:
\( 0 = a(0 - 3)^2 + 12 \)
\( 0 = 9a + 12 \)
Subtract 12 from both sides: \( -12 = 9a \)
Divide by 9: \( a = -\frac{12}{9} = -\frac{4}{3} \)

Step3: Write the final equation

Substitute \( a = -\frac{4}{3} \), \( h = 3 \), and \( k = 12 \) into the vertex form:
\( h(x) = -\frac{4}{3}(x - 3)^2 + 12 \)
We can also expand this to standard form:
\( h(x) = -\frac{4}{3}(x^2 - 6x + 9) + 12 \)
\( h(x) = -\frac{4}{3}x^2 + 8x - 12 + 12 \)
\( h(x) = -\frac{4}{3}x^2 + 8x \)

Answer:

The quadratic equation modeling the height is \( h(x) = -\frac{4}{3}(x - 3)^2 + 12 \) (or \( h(x) = -\frac{4}{3}x^2 + 8x \) in standard form).