Question
Which exponential function best models the data in the table?
x
−2
−1
0 1 2 3
y
1.75 3.5 7 14 28 56
Option #1: f(x)=(2⋅7)x
Option #2: f(x)=7⋅2x
Option #3: f(x)=2⋅7x
x
−2
−1
0 1 2 3
y
1.75 3.5 7 14 28 56
Option #1: f(x)=(2⋅7)x
Option #2: f(x)=7⋅2x
Option #3: f(x)=2⋅7x
Answers
Answered by
GPT-4o mini
To determine which exponential function best models the data in the table, we can analyze the values you've provided.
Let's consider the values of \( y \) at various \( x \):
\[
\begin{align*}
x &= -2, & y &= 1.75 \\
x &= -1, & y &= 3.5 \\
x &= 0, & y &= 7 \\
x &= 1, & y &= 14 \\
x &= 2, & y &= 28 \\
x &= 3, & y &= 56 \\
\end{align*}
\]
From the values \( y \) at \( x = 0 \) is \( 7 \) and appears to be a constant factor for each doubling of \( x \):
- From \( x = 0 \) to \( x = 1 \): \( 7 \to 14 \) (doubles)
- From \( x = 1 \) to \( x = 2 \): \( 14 \to 28 \) (doubles)
- From \( x = 2 \) to \( x = 3 \): \( 28 \to 56 \) (doubles)
- From \( x = -2 \) to \( x = -1 \): \( 1.75 \to 3.5 \) (doubles)
This indicates that \( y \) seems to double every time \( x \) is increased by \( 1 \), suggesting an exponential model of the form \( y = a \cdot b^x \) where \( b = 2 \).
Now let's evaluate the options provided:
1. **Option #1: \( f(x) = (2 \cdot 7)^x \)**
This simplifies to \( f(x) = 14^x \), which doesn't correspond to the pattern of doubling.
2. **Option #2: \( f(x) = 7 \cdot 2^x \)**
This form correctly identifies the base to be \( 2 \), and the initial value \( 7 \) matches our observation so far. This option should match our values perfectly:
- \( f(-2) = 7 \cdot 2^{-2} = 7 \cdot \frac{1}{4} = 1.75 \)
- \( f(-1) = 7 \cdot 2^{-1} = 7 \cdot \frac{1}{2} = 3.5 \)
- \( f(0) = 7 \cdot 2^0 = 7 \)
- \( f(1) = 7 \cdot 2^1 = 14 \)
- \( f(2) = 7 \cdot 2^2 = 28 \)
- \( f(3) = 7 \cdot 2^3 = 56 \)
3. **Option #3: \( f(x) = 2 \cdot 7^x \)**
This simplifies as \( f(x) = 2 \cdot 7^x \) and does not match the observed pattern of doubling.
Given this analysis, the best option that models the data is:
**Option #2: \( f(x) = 7 \cdot 2^x \)**
Let's consider the values of \( y \) at various \( x \):
\[
\begin{align*}
x &= -2, & y &= 1.75 \\
x &= -1, & y &= 3.5 \\
x &= 0, & y &= 7 \\
x &= 1, & y &= 14 \\
x &= 2, & y &= 28 \\
x &= 3, & y &= 56 \\
\end{align*}
\]
From the values \( y \) at \( x = 0 \) is \( 7 \) and appears to be a constant factor for each doubling of \( x \):
- From \( x = 0 \) to \( x = 1 \): \( 7 \to 14 \) (doubles)
- From \( x = 1 \) to \( x = 2 \): \( 14 \to 28 \) (doubles)
- From \( x = 2 \) to \( x = 3 \): \( 28 \to 56 \) (doubles)
- From \( x = -2 \) to \( x = -1 \): \( 1.75 \to 3.5 \) (doubles)
This indicates that \( y \) seems to double every time \( x \) is increased by \( 1 \), suggesting an exponential model of the form \( y = a \cdot b^x \) where \( b = 2 \).
Now let's evaluate the options provided:
1. **Option #1: \( f(x) = (2 \cdot 7)^x \)**
This simplifies to \( f(x) = 14^x \), which doesn't correspond to the pattern of doubling.
2. **Option #2: \( f(x) = 7 \cdot 2^x \)**
This form correctly identifies the base to be \( 2 \), and the initial value \( 7 \) matches our observation so far. This option should match our values perfectly:
- \( f(-2) = 7 \cdot 2^{-2} = 7 \cdot \frac{1}{4} = 1.75 \)
- \( f(-1) = 7 \cdot 2^{-1} = 7 \cdot \frac{1}{2} = 3.5 \)
- \( f(0) = 7 \cdot 2^0 = 7 \)
- \( f(1) = 7 \cdot 2^1 = 14 \)
- \( f(2) = 7 \cdot 2^2 = 28 \)
- \( f(3) = 7 \cdot 2^3 = 56 \)
3. **Option #3: \( f(x) = 2 \cdot 7^x \)**
This simplifies as \( f(x) = 2 \cdot 7^x \) and does not match the observed pattern of doubling.
Given this analysis, the best option that models the data is:
**Option #2: \( f(x) = 7 \cdot 2^x \)**