Answer :
Let's break down the steps to complete the table and find the equation of the exponential function [tex]\( y = a \cdot b^x \)[/tex].
Step-by-Step Solution:
### Part a: Completing the Table
Given:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & \square \\ \hline 1 & 512 \\ \hline 2 & 128 \\ \hline 3 & 32 \\ \hline 4 & \square \\ \hline \end{array} \][/tex]
1. Identify the base [tex]\(b\)[/tex]:
- We observe that [tex]\( y \)[/tex] changes as [tex]\( x \)[/tex] increments. To find [tex]\( b \)[/tex], we can take the ratio of consecutive [tex]\( y \)[/tex]-values.
- Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ b = \frac{y_2}{y_1} = \frac{128}{512} = \frac{1}{4} = 0.25 \][/tex]
2. Determine the coefficient [tex]\(a\)[/tex]:
- Using the known pair [tex]\((1, 512)\)[/tex]:
[tex]\[ 512 = a \cdot (0.25)^1 \implies a = 512 \cdot \frac{1}{0.25} = 512 \cdot 4 = 2048 \][/tex]
3. Calculate missing [tex]\(y\)[/tex]-values:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y_0 = a \cdot (0.25)^0 = 2048 \cdot 1 = 2048 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ y_4 = a \cdot (0.25)^4 = 2048 \cdot \left(\frac{1}{256}\right) = \frac{2048}{256} = 8 \][/tex]
The completed table is:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 2048.0 \\ \hline 1 & 512 \\ \hline 2 & 128 \\ \hline 3 & 32 \\ \hline 4 & 8.0 \\ \hline \end{array} \][/tex]
### Part b: Equation of the Exponential Function
Given the form [tex]\( y = a \cdot b^x \)[/tex], we have:
- Coefficient [tex]\( a \)[/tex]:
[tex]\[ a = 2048.0 \][/tex]
- Base [tex]\( b \)[/tex]:
[tex]\[ b = 0.25 \][/tex]
To summarize:
- Completed values in the table for [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex] and [tex]\( x = 4 \)[/tex]:
- [tex]\( y \)[/tex] for [tex]\( x = 0 \)[/tex] is 2048.0
- [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex] is 8.0
- The values for the equation are:
[tex]\[ a = 2048.0, \quad b = 0.25 \][/tex]
Step-by-Step Solution:
### Part a: Completing the Table
Given:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & \square \\ \hline 1 & 512 \\ \hline 2 & 128 \\ \hline 3 & 32 \\ \hline 4 & \square \\ \hline \end{array} \][/tex]
1. Identify the base [tex]\(b\)[/tex]:
- We observe that [tex]\( y \)[/tex] changes as [tex]\( x \)[/tex] increments. To find [tex]\( b \)[/tex], we can take the ratio of consecutive [tex]\( y \)[/tex]-values.
- Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ b = \frac{y_2}{y_1} = \frac{128}{512} = \frac{1}{4} = 0.25 \][/tex]
2. Determine the coefficient [tex]\(a\)[/tex]:
- Using the known pair [tex]\((1, 512)\)[/tex]:
[tex]\[ 512 = a \cdot (0.25)^1 \implies a = 512 \cdot \frac{1}{0.25} = 512 \cdot 4 = 2048 \][/tex]
3. Calculate missing [tex]\(y\)[/tex]-values:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y_0 = a \cdot (0.25)^0 = 2048 \cdot 1 = 2048 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ y_4 = a \cdot (0.25)^4 = 2048 \cdot \left(\frac{1}{256}\right) = \frac{2048}{256} = 8 \][/tex]
The completed table is:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 2048.0 \\ \hline 1 & 512 \\ \hline 2 & 128 \\ \hline 3 & 32 \\ \hline 4 & 8.0 \\ \hline \end{array} \][/tex]
### Part b: Equation of the Exponential Function
Given the form [tex]\( y = a \cdot b^x \)[/tex], we have:
- Coefficient [tex]\( a \)[/tex]:
[tex]\[ a = 2048.0 \][/tex]
- Base [tex]\( b \)[/tex]:
[tex]\[ b = 0.25 \][/tex]
To summarize:
- Completed values in the table for [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex] and [tex]\( x = 4 \)[/tex]:
- [tex]\( y \)[/tex] for [tex]\( x = 0 \)[/tex] is 2048.0
- [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex] is 8.0
- The values for the equation are:
[tex]\[ a = 2048.0, \quad b = 0.25 \][/tex]
