Question 1
Part A: During what interval(s) of the domain is the water balloon's height increasing?
The water balloon's height is increasing between 0 seconds and 2 seconds. In this interval, the height increases from 60 feet (at x = 0) to 75 feet (at x = 2).
Part B: During what interval(s) of the domain is the water balloon's height staying the same?
The water balloon's height remains the same from 2 seconds until 4 seconds. The height is constant at 75 feet during this interval.
Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest?
The water balloon's height decreases the fastest between 4 seconds and 6 seconds, where it drops from 75 feet down to 40 feet. This is a decrease of 35 feet over 2 seconds, which corresponds to a vertical drop of 17.5 feet per second.
Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds.
By the time we reach 12 seconds, the water balloon has already reached a height of 0 feet, meaning it has hit the ground. Therefore, the height of the water balloon at 16 seconds will also be 0 feet, as it cannot be below ground level.
Question 2
Part A: Determine the test average for your math class after completing test 2.
To find the average after test 2, substitute x = 2 into the mathematical function:
\[ f(2) = 0.9(2) + 79 = 1.8 + 79 = 80.8 \]
So, the average score in your math class after test 2 is 80.8.
Part B: Determine the test average for your science class after completing test 2.
From the provided table for the science class:
The average after test 2 is 79.
Part C: Which class had a higher average after completing test 4? Show work to support your answer.
First, calculate the average score for the math class after test 4:
\[ f(4) = 0.9(4) + 79 = 3.6 + 79 = 82.6 \]
Next, from the provided values for the science class, the average score at test 4 is not provided directly. We must assume a trend or interpolation. If we assume consistent increments based on the test scores:
- Test 1: 78
- Test 2: 79
- Test 3: 80,
We notice an increment of 1 point for each test.
Thus, test 4's average for science:
\[ g(4) = 80 + 1 = 81. \]
Now compare:
- Math Average after Test 4: 82.6
- Science Average after Test 4: 81
The math class had a higher average after completing test 4 at 82.6 compared to the science class at 81.
Question 3
Part A: Find and interpret the slope of the function.
To find the slope \( m \) of the function g(x) between any two points, we can use the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Taking points (0, 1500) and (2, 1350):
\[ m = \frac{1350 - 1500}{2 - 0} = \frac{-150}{2} = -75 \]
The slope of -75 indicates that for each day that passes, the bank account balance decreases by $75.
Part B: Write the equation of the line in point-slope, slope-intercept, and standard forms.
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Point-Slope Form: Using the point (0, 1500): \[ g(x) - 1500 = -75(x - 0) \rightarrow g(x) = -75x + 1500 \]
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Slope-Intercept Form: \[ g(x) = -75x + 1500 \]
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Standard Form: Rearranging: \[ 75x + g(x) = 1500 \] It can also be written as: \[ 75x + y = 1500 \]
Part C: Write the equation of the line using function notation.
Using function notation: \[ g(x) = -75x + 1500 \]
Part D: What is the balance in the bank account after 5 days?
To find the balance after 5 days, substitute x = 5 into g(x):
\[ g(5) = -75(5) + 1500 = -375 + 1500 = 1125 \]
The balance in the bank account after 5 days is $1125.
Question 4
Part A: Given the function g(x) = |x − 5|, describe the graph of the function, including the vertex, domain, and range.
The graph of \( g(x) = |x - 5| \) is a V-shape opening upwards.
- Vertex: The vertex is at the point (5, 0) since this is where the function value is minimized.
- Domain: The domain is all real numbers, \( (-\infty, \infty) \).
- Range: The range is \( [0, \infty) \) because the absolute value function cannot be below zero.
Part B: If the parent function f(x) = |x| is transformed to h(x) = |x| + 3, what transformation occurs from f(x) to h(x)? How are the vertex and range of h(x) affected?
The transformation from \( f(x) = |x| \) to \( h(x) = |x| + 3 \) is a vertical shift upward by 3 units.
- Vertex: The vertex of f(x) = |x| is at (0, 0). Therefore, the vertex of h(x) = |x| + 3 moves up to (0, 3).
- Range: The range of the parent function \( f(x) \) is \( [0, \infty) \). The transformation shifts it to \( [3, \infty) \) for h(x).
This indicates that the minimum value is now 3 due to the upward shift.