Question
which type of function best models the data in the table?
x: 2, 5, 8
y: 0.5, -1,75, -4
a linear
b quadratic
c exponential
d cubic
x: 2, 5, 8
y: 0.5, -1,75, -4
a linear
b quadratic
c exponential
d cubic
Answers
Answered by
GPT-4o mini
To determine which type of function best models the given data points, we can first examine the provided values in the table:
- For \(x = 2\), \(y = 0.5\)
- For \(x = 5\), \(y = -1.75\)
- For \(x = 8\), \(y = -4\)
### Step 1: Calculate the differences in \(y\) values
1. \(y(2) = 0.5\)
2. \(y(5) = -1.75\)
3. \(y(8) = -4\)
Now, let's look at the differences:
- From \(y(2)\) to \(y(5)\):
\[
-1.75 - 0.5 = -2.25
\]
- From \(y(5)\) to \(y(8)\):
\[
-4 - (-1.75) = -2.25
\]
### Step 2: Check if there are constant second differences
Now let's calculate the second differences:
1. The first differences are both \(-2.25\), which is constant.
Since the first differences show a constant value, we check the second differences:
- The first differences will yield only one unique value, thus not giving a second difference.
### Conclusion
Since we can observe that the changes in the \(y\) values seem to decrease linearly, and there is no constant rate of change in the second differences, we can rule out quadratic and cubic.
**The best fit for the given data is a linear function.**
**Answer: A. Linear**
- For \(x = 2\), \(y = 0.5\)
- For \(x = 5\), \(y = -1.75\)
- For \(x = 8\), \(y = -4\)
### Step 1: Calculate the differences in \(y\) values
1. \(y(2) = 0.5\)
2. \(y(5) = -1.75\)
3. \(y(8) = -4\)
Now, let's look at the differences:
- From \(y(2)\) to \(y(5)\):
\[
-1.75 - 0.5 = -2.25
\]
- From \(y(5)\) to \(y(8)\):
\[
-4 - (-1.75) = -2.25
\]
### Step 2: Check if there are constant second differences
Now let's calculate the second differences:
1. The first differences are both \(-2.25\), which is constant.
Since the first differences show a constant value, we check the second differences:
- The first differences will yield only one unique value, thus not giving a second difference.
### Conclusion
Since we can observe that the changes in the \(y\) values seem to decrease linearly, and there is no constant rate of change in the second differences, we can rule out quadratic and cubic.
**The best fit for the given data is a linear function.**
**Answer: A. Linear**
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