Question
A water balloon is launched up from the ground. After 2.4 seconds, the balloon reaches a maximum height of 28.8 meters.
It returns to the ground 4.8 seconds after it is launched.
Let / (x) be the height (in meters) of the balloon x seconds after it is launched. Then, the function / is guadratic. (Its graph is a parabola.) Write an equation for the quadratic function f.
It returns to the ground 4.8 seconds after it is launched.
Let / (x) be the height (in meters) of the balloon x seconds after it is launched. Then, the function / is guadratic. (Its graph is a parabola.) Write an equation for the quadratic function f.
Answers
GPT-4o mini
To write the quadratic function \( f(x) \) representing the height of the water balloon, we can use the vertex form of a quadratic equation:
\[
f(x) = a(x - h)^2 + k
\]
where \( (h, k) \) is the vertex of the parabola. From the information given:
- The maximum height of the balloon is \( 28.8 \) meters, which occurs at \( x = 2.4 \) seconds. So, the vertex is \( (h, k) = (2.4, 28.8) \).
- The balloon returns to the ground after \( 4.8 \) seconds, which means \( f(4.8) = 0 \).
Since the balloon is launched from the ground, we can also assume \( f(0) = 0 \) (as it starts at height 0).
We can plug the vertex into the vertex form equation:
\[
f(x) = a(x - 2.4)^2 + 28.8
\]
Next, we need to find the value of \( a \). We can use the information that \( f(0) = 0 \):
\[
0 = a(0 - 2.4)^2 + 28.8
\]
Solving for \( a \):
\[
0 = a(5.76) + 28.8
\]
\[
a(5.76) = -28.8
\]
\[
a = \frac{-28.8}{5.76} = -5
\]
Now we have \( a = -5 \). So the equation for the quadratic function is:
\[
f(x) = -5(x - 2.4)^2 + 28.8
\]
Now we can verify if this function returns to the ground at \( x = 4.8 \):
\[
f(4.8) = -5(4.8 - 2.4)^2 + 28.8
\]
\[
= -5(2.4)^2 + 28.8
\]
\[
= -5(5.76) + 28.8
\]
\[
= -28.8 + 28.8 = 0
\]
Thus, the quadratic function for the height of the water balloon is:
\[
f(x) = -5(x - 2.4)^2 + 28.8
\]
\[
f(x) = a(x - h)^2 + k
\]
where \( (h, k) \) is the vertex of the parabola. From the information given:
- The maximum height of the balloon is \( 28.8 \) meters, which occurs at \( x = 2.4 \) seconds. So, the vertex is \( (h, k) = (2.4, 28.8) \).
- The balloon returns to the ground after \( 4.8 \) seconds, which means \( f(4.8) = 0 \).
Since the balloon is launched from the ground, we can also assume \( f(0) = 0 \) (as it starts at height 0).
We can plug the vertex into the vertex form equation:
\[
f(x) = a(x - 2.4)^2 + 28.8
\]
Next, we need to find the value of \( a \). We can use the information that \( f(0) = 0 \):
\[
0 = a(0 - 2.4)^2 + 28.8
\]
Solving for \( a \):
\[
0 = a(5.76) + 28.8
\]
\[
a(5.76) = -28.8
\]
\[
a = \frac{-28.8}{5.76} = -5
\]
Now we have \( a = -5 \). So the equation for the quadratic function is:
\[
f(x) = -5(x - 2.4)^2 + 28.8
\]
Now we can verify if this function returns to the ground at \( x = 4.8 \):
\[
f(4.8) = -5(4.8 - 2.4)^2 + 28.8
\]
\[
= -5(2.4)^2 + 28.8
\]
\[
= -5(5.76) + 28.8
\]
\[
= -28.8 + 28.8 = 0
\]
Thus, the quadratic function for the height of the water balloon is:
\[
f(x) = -5(x - 2.4)^2 + 28.8
\]