To address each part of the question, I will create two hypothetical graphs based on the given points, derive the corresponding equations, and develop real-world situations for both.
Graph 1: Points (-1, 11) and (1, 0.44)
A. Input-Output Table
| x | y | |----|---------| | -1 | 11 | | 0 | 5.5 | | 1 | 0.44 | | 2 | 0.24 |
B. Exponential Equation
To find the exponential equation, we can assume a model of the form \( y = a \cdot b^x \).
- Starting with the known points:
- From point (-1, 11): \( 11 = a \cdot b^{-1} \) → \( 11b = a \) (1)
- From point (1, 0.44): \( 0.44 = a \cdot b^1 \) → \( a \cdot b = 0.44 \) (2)
From equations (1) and (2), we can substitute \( a = 11b \) into (2): \( 11b \cdot b = 0.44 \) → \( 11b^2 = 0.44 \) → \( b^2 = \frac{0.44}{11} = 0.04 \) → \( b = \sqrt{0.04} = 0.2 \)
Substituting \( b = 0.2 \) back into (1): \( a = 11(0.2) = 2.2 \)
Thus, the equation is: \[ y = 2.2 \cdot (0.2)^x \]
C. Real-World Situation
A possible scenario could be that this graph models the decrease in the number of certain species in an isolated environment due to competition over time. At \( x = -1 \) (representing a time before observation), there were 11 species observed initially.
D. Explanation of Variables
- \( x \): Time (in years)
- \( y \): Number of species present
- \( a \): Initial number of species present (2.2 species would not be practically meaningful but is the modeled value)
- \( b \): Growth/decay rate (0.2, indicating a decrease in number)
E. Evaluation
Evaluate for \( x = 3\): \[ y = 2.2 \cdot (0.2)^3 = 2.2 \cdot 0.008 = 0.0176 \]
This indicates that after 3 years, the projected number of species might be approximately 0.0176, signifying almost extinction.
Graph 2: Points (-1, 2.5) and (2, 1.258)
A. Input-Output Table
| x | y | |----|---------| | -1 | 2.5 | | 0 | 2.0 | | 1 | 1.6 | | 2 | 1.258 |
B. Exponential Equation
Using the exponential form again, \( y = a \cdot b^x \):
- From point (-1, 2.5): \( 2.5 = a \cdot b^{-1} \) → \( 2.5b = a \) (1)
- From point (0, 2.0): \( 2.0 = a \cdot b^0 \) → \( a = 2.0 \) (2)
Substituting (2) into (1):
- \( 2.5b = 2.0 \) → \( b = \frac{2.0}{2.5} = 0.8 \)
Thus, the equation is: \[ y = 2.0 \cdot (0.8)^x \]
C. Real-World Situation
This graph could represent the rate of depreciation of a car value over time. At \( x = -1 \) (after purchase), the car was initially valued at 2.5 times a particular base value.
D. Explanation of Variables
- \( x \): Time in years after purchase
- \( y \): Value of the car
- \( a \): Initial value of the car (2.0, could represent a value like $20,000)
- \( b \): Decay rate (0.8, meaning the car retains 80% of its value each year)
E. Evaluation
Evaluate for \( x = 3\): \[ y = 2.0 \cdot (0.8)^3 = 2.0 \cdot 0.512 = 1.024 \]
This indicates that after 3 years, the projected value of the car could be approximately 1.024 times the base value.
F. Unique Exponential Decay Situation
Let's create an example modeling the decay of a radioactive substance.
Scenario:
A radioactive element has an initial quantity of 100 grams, and it decays at a rate of 0.5 each year.
Equation:
\[ y = 100 \cdot (0.5)^x \]
Input-Output Table
| x | y | |----|---------| | 0 | 100 | | 1 | 50 | | 2 | 25 | | 3 | 12.5 |
Graph:
This would be a typical exponential decay curve beginning at 100 and halving each year.
Explanation of Variables:
- \( x \): Time in years
- \( y \): Quantity of the substance remaining
- \( a \): Initial quantity (100 grams)
- \( b \): Decay rate (0.5, representing a decay to half the quantity each year)
By evaluating, for example, when \( x = 4 \): \[ y = 100 \cdot (0.5)^4 = 100 \cdot 0.0625 = 6.25 \] This implies 6.25 grams of the original substance remains after 4 years.