To determine the function that models the given data, we will analyze the pattern in the values of [tex]\( y \)[/tex] as [tex]\( x \)[/tex] changes.
Here is the provided table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-1 & 1 \\
\hline
0 & 2 \\
\hline
1 & 4 \\
\hline
2 & 8 \\
\hline
3 & 16 \\
\hline
\end{array}
\][/tex]
We need to decide whether a linear, quadratic, or exponential function best fits the data.
1. Linear function ([tex]\( y = mx + b \)[/tex]):
- Calculate the differences in [tex]\( y \)[/tex] values:
1. [tex]\( 2 - 1 = 1 \)[/tex]
2. [tex]\( 4 - 2 = 2 \)[/tex]
3. [tex]\( 8 - 4 = 4 \)[/tex]
4. [tex]\( 16 - 8 = 8 \)[/tex]
The differences are not constant, so the data is not linear.
2. Quadratic function ([tex]\( y = ax^2 \)[/tex]):
- Calculate the second differences, which are the differences of the first differences:
1. [tex]\( 2 - 1 = 1 \)[/tex]
2. [tex]\( 4 - 2 = 2 \)[/tex]
3. [tex]\( 8 - 4 = 4 \)[/tex]
4. [tex]\( 16 - 8 = 8 \)[/tex]
The second differences are also not constant. Thus, the data does not fit a quadratic model.
3. Exponential function ([tex]\( y = a \cdot b^x \)[/tex]):
- Analyze the ratio between consecutive [tex]\( y \)[/tex] values:
1. [tex]\( \frac{2}{1} = 2 \)[/tex]
2. [tex]\( \frac{4}{2} = 2 \)[/tex]
3. [tex]\( \frac{8}{4} = 2 \)[/tex]
4. [tex]\( \frac{16}{8} = 2 \)[/tex]
The ratio between consecutive values is constant (2), indicating an exponential relationship.
We will construct an exponential function in the form [tex]\( y = a \cdot b^x \)[/tex]:
- From [tex]\( x = 0 \)[/tex] and [tex]\( y = 2 \)[/tex], we can see that:
[tex]\[
2 = a \cdot b^0 \implies 2 = a \cdot 1 \implies a = 2
\][/tex]
- Next, we use the point [tex]\( (x = 1, y = 4) \)[/tex]:
[tex]\[
4 = 2 \cdot b^1 \implies 4 = 2 \cdot b \implies b = 2
\][/tex]
Thus, the exponential function that models the data is:
[tex]\[ y = 2 \cdot (2^x) \][/tex]
So, the function is:
[tex]\[ y = 2 \cdot (2^x) \][/tex]
