Question

The function f(t) = t2 + 6t − 20 represents a parabola.

Part A: Rewrite the function in vertex form by completing the square. Show your work. (6 points)

Part B: Determine the vertex and indicate whether it is a maximum or a minimum on the graph. How do you know? (2 points)
The function H(t) = −16t2 + 90t + 50 shows the height H(t), in feet, of a projectile after t seconds. A second object moves in the air along a path represented by g(t) = 28 + 48.8t, where g(t) is the height, in feet, of the object from the ground at time t seconds.

Part A: Create a table using integers 1 through 4 for the 2 functions. Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? (6 points)

Part B: Explain what the solution from Part A means in the context of the problem. (4 points)

Answer 1

The correct answers are:

Question 1 - Part A: f(t)=(t+3)²-29; Part B: (-3, -29), minimum; Question 2 - Part A: H(1) = 124, g(1) = 76.8; H(2) = 166, g(2) = 125.6; H(3) = 176, g(3) = 174.4; H(4) = 154, g(4) = 223.2; Part B: Between 3 and 4 seconds, because that is where the values of g(t) catch up with H(t).

Step-by-step explanation:

Our quadratic function is in the form f(x)=ax²+bx+c. Our value of a is 1, b is 6, and c is -20.

To write a quadratic in vertex form, first take half of the b value and square it: (6/2)² = 3² = 9. This is what we will add and subtract to the function:

f(t) = t²+6t+9-20-9

The squared portion will be (t+b/2)²:

f(t) = (t+3)²-20-9

f(t) = (t+3)²-29

Vertex form is f(x) = a(x-h)²+k, where (h, k) is the vertex; in our function, (h, k) is (-3, -29).

Since the value of a was a positive, this parabola opens upward; this makes the vertex a minimum.

For Question 2 Part A, substitute the values 1, 2, 3 and 4 in H(t) and g(t).

For Part B, we can see that the values of g(t) are much less than that of H(t) until 3 seconds. From there, we can see that g(t) passes H(t). This means that the solution point, where they intersect, is between 3 and 4 seconds.

Answer 2

Part A: f(t) = t² + 6t - 20
u = t² + 6t - 20
+ 20 + 20
u + 20 = t² + 6t
u + 20 + 9 = t² + 6t + 9
u + 29 = t² + 3t + 3t + 9
u + 29 = t(t) + t(3) + 3(t) + 3(3)
u + 29 = t(t + 3) + 3(t + 3)
u + 29 = (t + 3)(t + 3)
u + 29 = (t + 3)²
- 29 - 29
u = (t + 3)² - 29

Part B: The vertex is (-3, -29). The graph shows that it is a minimum because it shows that there is a positive sign before the x²-term, making the parabola open up and has a minimum vertex of (-3, -29).
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Part A: g(t) = 48.8t + 28 h(t) = -16t² + 90t + 50
| t | g(t) | | t | h(t) |
|-4|-167.2| | -4 | -566 |
|-3|-118.4| | -3 | -364 |
|-2| -69.6 | | -2 | -194 |
|-1| -20.8 | | -1 | -56 |
|0 | -28 | | 0 | 50 |
|1 | 76.8 | | 1 | 124 |
|2 | 125.6| | 2 | 166 |
|3 | 174.4| | 3 | 176 |
|4 | 223.2| | 4 | 154 |
The two seconds that the solution of g(t) and h(t) is located is between -1 and 4 seconds because it shows that they have two solutions, making it between -1 and 4 seconds.

Part B: The solution from Part A means that you have to find two solutions in order to know where the solutions of the two functions are located at.