Answer :
To solve this problem, let's analyze the function [tex]\( f(x) = (0.5)^x \)[/tex], and fill in the table and additional required information accordingly.
### Step-by-step Solution:
1. Calculate the Function Values:
We need to determine the values of [tex]\( f(x) \)[/tex] for different [tex]\( x \)[/tex].
- For [tex]\( x = -2 \)[/tex]: [tex]\( f(-2) = (0.5)^{-2} = 4.0 \)[/tex]
- For [tex]\( x = -1 \)[/tex]: [tex]\( f(-1) = (0.5)^{-1} = 2.0 \)[/tex]
- For [tex]\( x = 0 \)[/tex]: [tex]\( f(0) = (0.5)^0 = 1.0 \)[/tex]
- For [tex]\( x = 1 \)[/tex]: [tex]\( f(1) = (0.5)^1 = 0.5 \)[/tex]
- For [tex]\( x = 2 \)[/tex]: [tex]\( f(2) = (0.5)^2 = 0.25 \)[/tex]
So, the table should be filled as follows:
[tex]\[ \begin{tabular}{|c|c|cc|c|c|} \hline $x$ & -2 & -1 & 0 & 1 & 2 \\ \hline $f(x)$ & 4.0 & 2.0 & 1.0 & 0.5 & 0.25 \\ \hline \end{tabular} \][/tex]
2. Identify Constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
From the hint, [tex]\( f(x) \)[/tex] can be written as [tex]\( a \left( \frac{1}{2} \right)^x \)[/tex].
- Here, [tex]\( a \)[/tex] is the initial value when [tex]\( x = -2 \)[/tex], so [tex]\( a = 4.0 \)[/tex].
- The base [tex]\( b \)[/tex] is given as [tex]\( 0.5 \)[/tex].
3. Determine the [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept is the value of [tex]\( f(x) \)[/tex] when [tex]\( x = 0 \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( f(0) = 1.0 \)[/tex].
- Therefore, [tex]\( y \)[/tex]-intercept is [tex]\( 1.0 \)[/tex].
4. Analyze the End Behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( (0.5)^x \)[/tex] approaches [tex]\( 0 \)[/tex] because any positive fraction raised to an increasingly large power approaches zero. Therefore:
[tex]\[ \text{As } x \rightarrow \infty, \; y \rightarrow 0. \][/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( (0.5)^x \)[/tex] approaches [tex]\( \infty \)[/tex] because the base [tex]\( 0.5 \)[/tex] raised to an increasingly large negative power (which is the same as raising [tex]\( 2 \)[/tex] to the corresponding positive power) increases without bound. Therefore:
[tex]\[ \text{As } x \rightarrow -\infty, \; y \rightarrow \infty. \][/tex]
Summarizing the results:
[tex]\[ \begin{tabular}{|c|c|cc|c|c|} \hline $x$ & -2 & -1 & 0 & 1 & 2 \\ \hline $f(x)$ & 4.0 & 2.0 & 1.0 & 0.5 & 0.25 \\ \hline \end{tabular} \][/tex]
Given:
[tex]\[ a = 4.0, \; b = 0.5, \; y\text{-intercept} = 1.0 \][/tex]
End Behavior:
[tex]\[ \text{As } x \rightarrow \infty, \; y \rightarrow 0 \][/tex]
[tex]\[ \text{As } x \rightarrow -\infty, \; y \rightarrow \infty \][/tex]
### Step-by-step Solution:
1. Calculate the Function Values:
We need to determine the values of [tex]\( f(x) \)[/tex] for different [tex]\( x \)[/tex].
- For [tex]\( x = -2 \)[/tex]: [tex]\( f(-2) = (0.5)^{-2} = 4.0 \)[/tex]
- For [tex]\( x = -1 \)[/tex]: [tex]\( f(-1) = (0.5)^{-1} = 2.0 \)[/tex]
- For [tex]\( x = 0 \)[/tex]: [tex]\( f(0) = (0.5)^0 = 1.0 \)[/tex]
- For [tex]\( x = 1 \)[/tex]: [tex]\( f(1) = (0.5)^1 = 0.5 \)[/tex]
- For [tex]\( x = 2 \)[/tex]: [tex]\( f(2) = (0.5)^2 = 0.25 \)[/tex]
So, the table should be filled as follows:
[tex]\[ \begin{tabular}{|c|c|cc|c|c|} \hline $x$ & -2 & -1 & 0 & 1 & 2 \\ \hline $f(x)$ & 4.0 & 2.0 & 1.0 & 0.5 & 0.25 \\ \hline \end{tabular} \][/tex]
2. Identify Constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
From the hint, [tex]\( f(x) \)[/tex] can be written as [tex]\( a \left( \frac{1}{2} \right)^x \)[/tex].
- Here, [tex]\( a \)[/tex] is the initial value when [tex]\( x = -2 \)[/tex], so [tex]\( a = 4.0 \)[/tex].
- The base [tex]\( b \)[/tex] is given as [tex]\( 0.5 \)[/tex].
3. Determine the [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept is the value of [tex]\( f(x) \)[/tex] when [tex]\( x = 0 \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( f(0) = 1.0 \)[/tex].
- Therefore, [tex]\( y \)[/tex]-intercept is [tex]\( 1.0 \)[/tex].
4. Analyze the End Behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( (0.5)^x \)[/tex] approaches [tex]\( 0 \)[/tex] because any positive fraction raised to an increasingly large power approaches zero. Therefore:
[tex]\[ \text{As } x \rightarrow \infty, \; y \rightarrow 0. \][/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( (0.5)^x \)[/tex] approaches [tex]\( \infty \)[/tex] because the base [tex]\( 0.5 \)[/tex] raised to an increasingly large negative power (which is the same as raising [tex]\( 2 \)[/tex] to the corresponding positive power) increases without bound. Therefore:
[tex]\[ \text{As } x \rightarrow -\infty, \; y \rightarrow \infty. \][/tex]
Summarizing the results:
[tex]\[ \begin{tabular}{|c|c|cc|c|c|} \hline $x$ & -2 & -1 & 0 & 1 & 2 \\ \hline $f(x)$ & 4.0 & 2.0 & 1.0 & 0.5 & 0.25 \\ \hline \end{tabular} \][/tex]
Given:
[tex]\[ a = 4.0, \; b = 0.5, \; y\text{-intercept} = 1.0 \][/tex]
End Behavior:
[tex]\[ \text{As } x \rightarrow \infty, \; y \rightarrow 0 \][/tex]
[tex]\[ \text{As } x \rightarrow -\infty, \; y \rightarrow \infty \][/tex]
