Question

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)=5⋅3^x
; f(−1)=5/3
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)=41.43x^2−73.71x+19.86
; f(−1)=135
f of x is equal to 41 point 4 3 x squared minus 73 point 7 1 x plus 19 point 8 6; f of negative 1 is equal to 135

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

Answer 1

To determine the function that best models the data and find the value of the function at \( x = -1 \), we first analyze the given data:

- \( f(0) = 5 \)
- \( f(1) = 15 \)
- \( f(2) = 45 \)
- \( f(3) = 135 \)
- \( f(4) = 405 \)

Looking at the y-values, we can see they seem to increase rapidly. This suggests a potential exponential relationship.

Let's evaluate the candidate functions one by one:

1. **Candidate: \( f(x) = 5 \cdot 3^x \)**
- \( f(0) = 5 \cdot 3^0 = 5 \)
- \( f(1) = 5 \cdot 3^1 = 15 \)
- \( f(2) = 5 \cdot 3^2 = 45 \)
- \( f(3) = 5 \cdot 3^3 = 135 \)
- \( f(4) = 5 \cdot 3^4 = 405 \)
- This function matches all given points.

To find \( f(-1) \):
\[
f(-1) = 5 \cdot 3^{-1} = 5 \cdot \frac{1}{3} = \frac{5}{3}
\]

2. **Candidate: \( f(x) = 41.43x^2 - 73.71x + 19.86 \)**
- You can evaluate this polynomial, but we noticed it doesn't seem to match the growth pattern.

3. **Candidate: \( f(x) = 2.23 \cdot (1.40)^x \)**
- Evaluating it yields values that don’t match the rapid growth.

4. **Candidate: \( f(x) = 92x - 63 \)**
- This is a linear function, which will not match the exponential growth either.

From the evaluation, the best model for the data is \( f(x) = 5 \cdot 3^x \).

Thus, the value of the function at \( x = -1 \) is:
\[
f(-1) = \frac{5}{3}
\]

**Final Answer:**
\( f(x) = 5 \cdot 3^x \); \( f(-1) = \frac{5}{3} \)