The path of a launched toy rocket can be modeled using a quadratic equation. Which of the models would be most helpful in finding the height at which the rocket was released?

A. [tex]y = -16 t(t-5) + 1[/tex]
B. [tex]y = -(t-3)^2 + 150[/tex]
C. [tex]y = -16(t+2)(t-8)[/tex]
D. [tex]y = -16 t^2 - 52 t + 3[/tex]



Answer :

To determine which model is the most helpful in finding the height at which the rocket was released, we need to evaluate each model at [tex]\( t = 0 \)[/tex], since that will give us the initial height of the rocket. Let's solve each equation to find the y-intercept, which corresponds to the height at [tex]\( t = 0 \)[/tex]:

1. Model: [tex]\( y = -16t(t-5) + 1 \)[/tex]
[tex]\[ y = -16 \cdot 0 \cdot (0 - 5) + 1 = 0 + 1 = 1 \][/tex]
So, the y-intercept is 1.

2. Model: [tex]\( y = -(t-3)^2 + 150 \)[/tex]
[tex]\[ y = -(0-3)^2 + 150 = -9 + 150 = 141 \][/tex]
So, the y-intercept is 141.

3. Model: [tex]\( y = -16(t+2)(t-8) \)[/tex]
[tex]\[ y = -16 \cdot (0+2) \cdot (0-8) = -16 \cdot 2 \cdot (-8) = -16 \cdot (-16) = 256 \][/tex]
So, the y-intercept is 256.

4. Model: [tex]\( y = -16t^2 - 52t + 3 \)[/tex]
[tex]\[ y = -16 \cdot 0^2 - 52 \cdot 0 + 3 = 0 + 0 + 3 = 3 \][/tex]
So, the y-intercept is 3.

Now, we have the y-intercepts for all four models:
- Model 1: 1
- Model 2: 141
- Model 3: 256
- Model 4: 3

The y-intercept represents the height at which the rocket was released. Therefore, the model with the y-intercept closest to the actual height of release would be the most helpful.

Based on the values computed, the height at which the rocket was released varies according to different models. The model giving the highest intercept, [tex]\( y = -16(t+2)(t-8) \)[/tex], which is 256, suggests it might capture a scenario depending on a more realistic initial height for the scenarios given in the problem context.