Asked by julian celeste
Use the table to answer the question.
x-Values y-Values
0 5
1 15
2 45
3 135
4 405
Select the function that best models the data and determine the value of the function at x=−1
.
(1 point)
Responses
f(x)=2.23⋅(1.40)x
; f(−1)=1.59
f of x is equal to 2 point 2 3 times 1 point 4 0 to the x th power ; f of negative 1 is equal to 1 point 5 9
f(x)=92x−63
; f(−1)=−155
f of x is equal to 92 x minus 63; f of negative 1 is equal to negative 155
f(x)=5⋅3x
; f(−1)=53
f of x is equal to 5 times 3 to the x th power; f of negative 1 is equal to 5 thirds
f(x)=41.43x2−73.71x+19.86
; f(−1)=135
x-Values y-Values
0 5
1 15
2 45
3 135
4 405
Select the function that best models the data and determine the value of the function at x=−1
.
(1 point)
Responses
f(x)=2.23⋅(1.40)x
; f(−1)=1.59
f of x is equal to 2 point 2 3 times 1 point 4 0 to the x th power ; f of negative 1 is equal to 1 point 5 9
f(x)=92x−63
; f(−1)=−155
f of x is equal to 92 x minus 63; f of negative 1 is equal to negative 155
f(x)=5⋅3x
; f(−1)=53
f of x is equal to 5 times 3 to the x th power; f of negative 1 is equal to 5 thirds
f(x)=41.43x2−73.71x+19.86
; f(−1)=135
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine which function best models the data provided in the table, we can analyze the given options and see which one fits the pattern of the y-values corresponding to the x-values.
The given table is:
\[
\begin{array}{|c|c|}
\hline
\text{x-Values} & \text{y-Values} \\
\hline
0 & 5 \\
1 & 15 \\
2 & 45 \\
3 & 135 \\
4 & 405 \\
\hline
\end{array}
\]
Upon quickly analyzing the changes in y-values as x increases, we notice that they are not linear. As x increases from 0 to 4, the y-values increase rapidly, suggesting an exponential or polynomial growth.
Let's evaluate the proposed functions one by one:
1. **f(x) = 2.23 × (1.40)^x; f(−1) = 1.59**
- This option looks like an exponential function. We would compute some values for x to check if it matches the data, but based on the rapid growth observed in the original data, it seems too low.
2. **f(x) = 92x − 63; f(−1) = −155**
- This appears to be a linear function. Plugging in x-values like 0, 1, 2, etc., we see it won't yield the correct y-values even before evaluating it at x=-1.
3. **f(x) = 5 × 3^x; f(−1) = 5/3**
- A potential exponential function. The y-values seem to be multiples of each other at successive x-values, and plugging in x=0 gives f(0) = 5, which is correct. Continuing, f(1) = 15, f(2) = 45, f(3) = 135, and f(4) = 405, all match the table. Therefore, this function fits perfectly.
4. **f(x) = 41.43x² − 73.71x + 19.86; f(−1) = 135**
- This appears to be a quadratic function. Evaluating it at x=0 gives f(0)=19.86, drastically different from the first data point, so it doesn't match.
The best model for the data is:
**f(x) = 5 × 3^x**.
Now, we need to calculate f(−1):
\[
f(−1) = 5 × 3^{-1} = 5 × \frac{1}{3} = \frac{5}{3}.
\]
So the answer is:
**f(x) = 5 × 3^x; f(−1) = \frac{5}{3}**.
The given table is:
\[
\begin{array}{|c|c|}
\hline
\text{x-Values} & \text{y-Values} \\
\hline
0 & 5 \\
1 & 15 \\
2 & 45 \\
3 & 135 \\
4 & 405 \\
\hline
\end{array}
\]
Upon quickly analyzing the changes in y-values as x increases, we notice that they are not linear. As x increases from 0 to 4, the y-values increase rapidly, suggesting an exponential or polynomial growth.
Let's evaluate the proposed functions one by one:
1. **f(x) = 2.23 × (1.40)^x; f(−1) = 1.59**
- This option looks like an exponential function. We would compute some values for x to check if it matches the data, but based on the rapid growth observed in the original data, it seems too low.
2. **f(x) = 92x − 63; f(−1) = −155**
- This appears to be a linear function. Plugging in x-values like 0, 1, 2, etc., we see it won't yield the correct y-values even before evaluating it at x=-1.
3. **f(x) = 5 × 3^x; f(−1) = 5/3**
- A potential exponential function. The y-values seem to be multiples of each other at successive x-values, and plugging in x=0 gives f(0) = 5, which is correct. Continuing, f(1) = 15, f(2) = 45, f(3) = 135, and f(4) = 405, all match the table. Therefore, this function fits perfectly.
4. **f(x) = 41.43x² − 73.71x + 19.86; f(−1) = 135**
- This appears to be a quadratic function. Evaluating it at x=0 gives f(0)=19.86, drastically different from the first data point, so it doesn't match.
The best model for the data is:
**f(x) = 5 × 3^x**.
Now, we need to calculate f(−1):
\[
f(−1) = 5 × 3^{-1} = 5 × \frac{1}{3} = \frac{5}{3}.
\]
So the answer is:
**f(x) = 5 × 3^x; f(−1) = \frac{5}{3}**.
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