Math Problem Statement

Write a linear (y = mx + b), quadratic (y = ax^2), or exponential (y = a(b)^x) function that models the data from the given table.

Solution

To analyze the data and determine whether a linear, quadratic, or exponential function models it best, let's first summarize the data from the table:

x & y \\ \hline 1 & \frac{7}{4} \\ 2 & \frac{7}{2} \\ 3 & 7 \\ 4 & 14 \\ 5 & 28 \\ \end{array}$$ We are given three possible models: 1. Linear: $$y = mx + b$$ 2. Quadratic: $$y = ax^2 + bx + c$$ 3. Exponential: $$y = a(b)^x$$ ### Step 1: Test for a Linear Model In a linear function, the difference between successive $$y$$-values should be constant. Let's calculate the differences: - $$y_2 - y_1 = \frac{7}{2} - \frac{7}{4} = \frac{7}{4}$$ - $$y_3 - y_2 = 7 - \frac{7}{2} = \frac{7}{2}$$ - $$y_4 - y_3 = 14 - 7 = 7$$ - $$y_5 - y_4 = 28 - 14 = 14$$ The differences are not constant, so a linear model does not fit. ### Step 2: Test for a Quadratic Model For a quadratic model, the second differences between successive $$y$$-values should be constant. We already have the first differences: - $$y_2 - y_1 = \frac{7}{4}$$ - $$y_3 - y_2 = \frac{7}{2}$$ - $$y_4 - y_3 = 7$$ - $$y_5 - y_4 = 14$$ Now, calculate the second differences: - $$\left( \frac{7}{2} - \frac{7}{4} \right) = \frac{7}{4}$$ - $$\left( 7 - \frac{7}{2} \right) = \frac{7}{2}$$ - $$\left( 14 - 7 \right) = 7$$ These are not constant either, so a quadratic model does not fit. ### Step 3: Test for an Exponential Model For an exponential function $$y = a(b)^x$$, the ratio between successive $$y$$-values should be constant. Let's calculate the ratios: - $$\frac{y_2}{y_1} = \frac{\frac{7}{2}}{\frac{7}{4}} = 2$$ - $$\frac{y_3}{y_2} = \frac{7}{\frac{7}{2}} = 2$$ - $$\frac{y_4}{y_3} = \frac{14}{7} = 2$$ - $$\frac{y_5}{y_4} = \frac{28}{14} = 2$$ Since the ratios are constant, the data fits an exponential model of the form $$y = a(b)^x$$. ### Step 4: Determine the Parameters for the Exponential Model We know the ratio $$b = 2$$. Now, let's determine $$a$$ by using the first data point $$(x = 1, y = \frac{7}{4})$$: $$\frac{7}{4} = a(2)^1 \implies a = \frac{7}{8}$$ Thus, the exponential model is: $$y = \frac{7}{8} \cdot 2^x$$ This equation models the given data. Would you like more details, or do you have any further questions? Here are five related questions to expand on this: 1. How would you solve this if given a different set of $$y$$-values? 2. Can you derive an exponential model using logarithms? 3. How would you identify a quadratic model if the second differences were constant? 4. Can you find the values of $$a$$ and $$b$$ if the ratio between successive terms was not an integer? 5. How would this process differ if the data followed a logarithmic pattern? **Tip**: Always check for patterns (constant differences or ratios) to identify whether a model is linear, quadratic, or exponential.

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Quadratic Equations
Exponential Functions
Sequences

Formulas

Linear: y = mx + b
Quadratic: y = ax^2 + bx + c
Exponential: y = a(b)^x

Theorems

None specified, but involves understanding differences and ratios for linearity, quadratic, and exponential growth.

Suitable Grade Level

Grades 9-12