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Certainly! Let’s analyze the exponential function [tex]\( f(x) = 5 \cdot 2^x \)[/tex].
### Step-by-Step Analysis
1. Identify [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
- The function is of the form [tex]\( f(x) = a \cdot b^x \)[/tex].
- Here, [tex]\(a = 5\)[/tex] and [tex]\(b = 2\)[/tex].
2. Calculate [tex]\(f(x)\)[/tex] for given [tex]\(x\)[/tex]-values
- We need to find [tex]\( f(x) \)[/tex] for [tex]\( x = -2, -1, 0, 1, \)[/tex] and [tex]\( 2 \)[/tex].
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 5 \cdot (2^{-2}) = 5 \cdot \frac{1}{4} = 1.25 \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 5 \cdot (2^{-1}) = 5 \cdot \frac{1}{2} = 2.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 5 \cdot (2^0) = 5 \cdot 1 = 5 \][/tex]
- This is also the value of the y-intercept.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 5 \cdot 2^1 = 5 \cdot 2 = 10 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 5 \cdot 2^2 = 5 \cdot 4 = 20 \][/tex]
3. Tabulate the values
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -2 & 1.25 \\ \hline -1 & 2.5 \\ \hline 0 & 5 \\ \hline 1 & 10 \\ \hline 2 & 20 \\ \hline \end{tabular} \][/tex]
4. Identify the y-intercept
- The y-intercept is [tex]\( f(0) \)[/tex]. From the calculation, [tex]\( f(0) = 5 \)[/tex]. Thus, the y-intercept is [tex]\( 5 \)[/tex].
5. Determine the end behavior
- As [tex]\( x \to +\infty \)[/tex], [tex]\( 2^x \)[/tex] grows without bound (because [tex]\( b > 1 \)[/tex]), so [tex]\( f(x) = 5 \cdot 2^x \to +\infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( 2^x \)[/tex] approaches 0 (since [tex]\( 0 < 2^x < 1 \)[/tex] when [tex]\( x \)[/tex] is negative), so [tex]\( f(x) = 5 \cdot 2^x \to 0 \)[/tex].
### Summary
- [tex]\( a = 5 \)[/tex]
- [tex]\( b = 2 \)[/tex]
- [tex]\( y \)[/tex]-intercept: [tex]\( 5 \)[/tex]
- End behavior:
- [tex]\( x \to +\infty \)[/tex]: [tex]\( f(x) \to +\infty \)[/tex]
- [tex]\( x \to -\infty \)[/tex]: [tex]\( f(x) \to 0 \)[/tex]
### Table of values
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -2 & 1.25 \\ \hline -1 & 2.5 \\ \hline 0 & 5 \\ \hline 1 & 10 \\ \hline 2 & 20 \\ \hline \end{tabular} \][/tex]
### Step-by-Step Analysis
1. Identify [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
- The function is of the form [tex]\( f(x) = a \cdot b^x \)[/tex].
- Here, [tex]\(a = 5\)[/tex] and [tex]\(b = 2\)[/tex].
2. Calculate [tex]\(f(x)\)[/tex] for given [tex]\(x\)[/tex]-values
- We need to find [tex]\( f(x) \)[/tex] for [tex]\( x = -2, -1, 0, 1, \)[/tex] and [tex]\( 2 \)[/tex].
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 5 \cdot (2^{-2}) = 5 \cdot \frac{1}{4} = 1.25 \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 5 \cdot (2^{-1}) = 5 \cdot \frac{1}{2} = 2.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 5 \cdot (2^0) = 5 \cdot 1 = 5 \][/tex]
- This is also the value of the y-intercept.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 5 \cdot 2^1 = 5 \cdot 2 = 10 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 5 \cdot 2^2 = 5 \cdot 4 = 20 \][/tex]
3. Tabulate the values
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -2 & 1.25 \\ \hline -1 & 2.5 \\ \hline 0 & 5 \\ \hline 1 & 10 \\ \hline 2 & 20 \\ \hline \end{tabular} \][/tex]
4. Identify the y-intercept
- The y-intercept is [tex]\( f(0) \)[/tex]. From the calculation, [tex]\( f(0) = 5 \)[/tex]. Thus, the y-intercept is [tex]\( 5 \)[/tex].
5. Determine the end behavior
- As [tex]\( x \to +\infty \)[/tex], [tex]\( 2^x \)[/tex] grows without bound (because [tex]\( b > 1 \)[/tex]), so [tex]\( f(x) = 5 \cdot 2^x \to +\infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( 2^x \)[/tex] approaches 0 (since [tex]\( 0 < 2^x < 1 \)[/tex] when [tex]\( x \)[/tex] is negative), so [tex]\( f(x) = 5 \cdot 2^x \to 0 \)[/tex].
### Summary
- [tex]\( a = 5 \)[/tex]
- [tex]\( b = 2 \)[/tex]
- [tex]\( y \)[/tex]-intercept: [tex]\( 5 \)[/tex]
- End behavior:
- [tex]\( x \to +\infty \)[/tex]: [tex]\( f(x) \to +\infty \)[/tex]
- [tex]\( x \to -\infty \)[/tex]: [tex]\( f(x) \to 0 \)[/tex]
### Table of values
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -2 & 1.25 \\ \hline -1 & 2.5 \\ \hline 0 & 5 \\ \hline 1 & 10 \\ \hline 2 & 20 \\ \hline \end{tabular} \][/tex]
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