Problem 1 - Algebra Form and Function

Question

Could the table represent the values of a linear function? $$ \begin{array}{c|c|c|c|c|c} \hline x & -6 & -3 & 0 & 3 & 6 \\ \hline y & 12 & 8 & 4 & 0 & -4 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

Answer: Yes, the table represents the values of a linear function because the relationship between the differences in x-values and y-values is constant.

Step 1: Find the differences between consecutive x and y values.

First, we will find the differences between consecutive x and y values: Difference between x-values: \((-3) - (-6) = 3\), \(0 - (-3) = 3\), \(3 - 0 = 3\), \(6 - 3 = 3\) Difference between y-values: \(8 - 12 = -4\), \(4 - 8 = -4\), \(0 - 4 = -4\), \((-4) - 0 = -4\) As we can see, the difference between consecutive x-values is 3, and the difference between consecutive y-values is -4.

Step 2: Check the relationship between x and y values.

Now, we will check if the relationship between the differences in x-values and y-values is constant: \(\frac{-4}{3}=\frac{-4}{3}=\frac{-4}{3}=\frac{-4}{3}\) We can see that the relationship between the differences in x-values and y-values is constant.

Step 3: Conclude if the function is linear or not.

Since the relationship between the differences in x-values and y-values is constant, the table represents the values of a linear function.

Key Concepts

Difference of X-Values

In any linear function, understanding the difference in x-values is crucial. The x-values in the table represent the input of the function, like independent variables. Calculating the difference between these values helps us understand the behaviour of the function’s rate of change. In this exercise, we have the x-values:

• -6
• -3
• 0
• 3
• 6

By subtracting each consecutive pair, we calculate:

• o\((-3) - (-6) = 3\)
• \(0 - (-3) = 3\)
• \(3 - 0 = 3\)
• \(6 - 3 = 3\)

As you can see, the difference between each pair is consistently 3. This tells us that the x-values are increasing by a fixed amount, which is the hallmark of a linear function. This kind of regularity in changes is a good hint at linearity, but we need to look further into the y-values to confirm it.

Difference of Y-Values

The difference in y-values acts like the change in output for each step up in the input. It shows how the function reacts to each change in x. Let's inspect how the y-values change:

• 12
• 8
• 4
• 0
• -4

We need to find the difference for each pair of consecutive y-values:

• \(8 - 12 = -4\)
• \(4 - 8 = -4\)
• \(0 - 4 = -4\)
• \((-4) - 0 = -4\)

The differences are all -4. A consistent difference in y-values as we move through x-values further supports the notion of a linear function. This regular change implies that the differences in input and output are proportionally linked, which is a key property of linear functions. To ensure this is a linear function, the relationship between these differences should also be constant.

Constant Relationship

In a linear function, the relationship between the difference of y-values and the difference of x-values must be constant. This relationship is often called the "slope" of the linear function, defined as the ratio of change in y to change in x.
Let's examine this with our calculated differences:Given the consistent differences:

• X-values difference = 3
• Y-values difference = -4

The slope can be computed as:

• \(\text{slope} = \frac{-4}{3}\)

Let's evaluate this for each segment of our table:

• Between -\((-6, 12)\) and \((-3, 8)\), the slope is \(\frac{-4}{3}\)
• Between \((-3, 8)\) and \((0, 4)\), the slope remains \(\frac{-4}{3}\)
• Similarly \((0, 4)\) to \((3, 0)\), it's \(\frac{-4}{3}\)
• And also between \((3, 0)\) to \((6, -4)\), \(\frac{-4}{3}\)

Because this slope remains constant across all intervals, the relationship is maintained. This tells us our function is indeed linear. With a consistent slope of \(\frac{-4}{3}\), each step in x results in a proportional change in y, confirming the linearity of our function.