To determine the equation of the quadratic function that models the height [tex]\( h(t) \)[/tex] of the softball at time [tex]\( t \)[/tex], we follow these steps:
### Step-by-Step Solution:
1. Identify Given Points and Key Characteristics:
- We know the height of the softball at three different points in time:
- After 1 second, [tex]\( h(1) = 30 \)[/tex] feet.
- After 3 seconds, [tex]\( h(3) = 38 \)[/tex] feet (Maximum height).
- After 5 seconds, [tex]\( h(5) = 30 \)[/tex] feet.
2. Recognize the Vertex Form of a Quadratic Function:
- A quadratic function in vertex form is given by:
[tex]\[
h(t) = a(t-h)^2 + k
\][/tex]
where [tex]\( (h, k) \)[/tex] is the vertex of the parabola.
3. Set the Vertex of the Parabola:
- Since the maximum height of 38 feet is achieved at [tex]\( t = 3 \)[/tex], the vertex is [tex]\( (3, 38) \)[/tex]. Thus, the quadratic function can be expressed as:
[tex]\[
h(t) = a(t-3)^2 + 38
\][/tex]
where [tex]\( a \)[/tex] is a coefficient to be determined.
4. Use the Additional Points to Determine [tex]\( a \)[/tex]:
- We use the point [tex]\( (1, 30) \)[/tex] to find [tex]\( a \)[/tex]:
[tex]\[
30 = a(1-3)^2 + 38
\][/tex]
Simplifying this, we get:
[tex]\[
30 = a(4) + 38
\][/tex]
[tex]\[
30 = 4a + 38
\][/tex]
[tex]\[
30 - 38 = 4a
\][/tex]
[tex]\[
-8 = 4a
\][/tex]
[tex]\[
a = -2
\][/tex]
5. Write the Final Equation:
- Substituting [tex]\( a = -2 \)[/tex] back into the vertex form, we get:
[tex]\[
h(t) = -2(t-3)^2 + 38
\][/tex]
6. Verify with the Other Points:
- Let’s check if [tex]\( h(5) = 30 \)[/tex] to confirm our equation:
[tex]\[
h(5) = -2(5-3)^2 + 38
\][/tex]
Simplifying this, we get:
[tex]\[
h(5) = -2(2)^2 + 38
\][/tex]
[tex]\[
h(5) = -2(4) + 38
\][/tex]
[tex]\[
h(5) = -8 + 38
\][/tex]
[tex]\[
h(5) = 30
\][/tex]
- The point [tex]\( (5, 30) \)[/tex] checks out as well.
Therefore, the equation of the quadratic function that models the height of the ball is:
[tex]\[
h(t) = -2(t-3)^2 + 38
\][/tex]
