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Sure! Let's analyze the exponential function [tex]\( f(x) = -3(3^x) \)[/tex] step-by-step:
### 1. Identify [tex]\( a \)[/tex] and [tex]\( b \)[/tex]
The general form of an exponential function is [tex]\( f(x) = a(b^x) \)[/tex]. Here, we can directly see:
- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 3 \)[/tex]
### 2. Determine the [tex]\( y \)[/tex]-intercept
The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex]. To find this, substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = -3 \cdot (3^0) = -3 \cdot 1 = -3 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is [tex]\(-3\)[/tex].
### 3. Calculate [tex]\( f(x) \)[/tex] for specified [tex]\( x \)[/tex] values
We need to find the values of [tex]\( f(x) \)[/tex] for [tex]\( x = -2, -1, 0, 1, 2 \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = -3 \cdot (3^{-2}) = -3 \cdot \frac{1}{3^2} = -3 \cdot \frac{1}{9} = -\frac{3}{9} = -0.3333 \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -3 \cdot (3^{-1}) = -3 \cdot \frac{1}{3} = -3 \cdot 0.3333 = -1 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -3 \cdot (3^0) = -3 \cdot 1 = -3 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -3 \cdot (3^1) = -3 \cdot 3 = -9 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = -3 \cdot (3^2) = -3 \cdot 9 = -27 \][/tex]
### 4. Determine the end behavior
Evaluate the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches positive and negative infinity:
- As [tex]\( x \to +\infty \)[/tex]: Since [tex]\( b = 3 > 1 \)[/tex] and the coefficient [tex]\( a = -3 \)[/tex] is negative, the function grows exponentially more negative. Graphically, this means the values of [tex]\( f(x) \)[/tex] become very large negative numbers.
So, as [tex]\( x \to +\infty \)[/tex], [tex]\( y \to -\infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex]: For very large negative values of [tex]\( x \)[/tex], the term [tex]\( 3^x \)[/tex] approaches 0 because any number raised to a negative power gets closer to zero. However, multiplied by [tex]\(-3\)[/tex], it becomes a very small negative number but still close to zero.
So, as [tex]\( x \to -\infty \)[/tex], [tex]\( y \to 0 \)[/tex].
### Summary
Here is the result:
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -0.3333 & -1 & -3 & -9 & -27 \\ \hline \end{array} \][/tex]
- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( y \)[/tex]-intercept = [tex]\(-3\)[/tex]
- End Behavior:
- As [tex]\( x \to +\infty \)[/tex], [tex]\( y \to -\infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( y \to 0 \)[/tex]
### 1. Identify [tex]\( a \)[/tex] and [tex]\( b \)[/tex]
The general form of an exponential function is [tex]\( f(x) = a(b^x) \)[/tex]. Here, we can directly see:
- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 3 \)[/tex]
### 2. Determine the [tex]\( y \)[/tex]-intercept
The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex]. To find this, substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = -3 \cdot (3^0) = -3 \cdot 1 = -3 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is [tex]\(-3\)[/tex].
### 3. Calculate [tex]\( f(x) \)[/tex] for specified [tex]\( x \)[/tex] values
We need to find the values of [tex]\( f(x) \)[/tex] for [tex]\( x = -2, -1, 0, 1, 2 \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = -3 \cdot (3^{-2}) = -3 \cdot \frac{1}{3^2} = -3 \cdot \frac{1}{9} = -\frac{3}{9} = -0.3333 \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -3 \cdot (3^{-1}) = -3 \cdot \frac{1}{3} = -3 \cdot 0.3333 = -1 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -3 \cdot (3^0) = -3 \cdot 1 = -3 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -3 \cdot (3^1) = -3 \cdot 3 = -9 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = -3 \cdot (3^2) = -3 \cdot 9 = -27 \][/tex]
### 4. Determine the end behavior
Evaluate the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches positive and negative infinity:
- As [tex]\( x \to +\infty \)[/tex]: Since [tex]\( b = 3 > 1 \)[/tex] and the coefficient [tex]\( a = -3 \)[/tex] is negative, the function grows exponentially more negative. Graphically, this means the values of [tex]\( f(x) \)[/tex] become very large negative numbers.
So, as [tex]\( x \to +\infty \)[/tex], [tex]\( y \to -\infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex]: For very large negative values of [tex]\( x \)[/tex], the term [tex]\( 3^x \)[/tex] approaches 0 because any number raised to a negative power gets closer to zero. However, multiplied by [tex]\(-3\)[/tex], it becomes a very small negative number but still close to zero.
So, as [tex]\( x \to -\infty \)[/tex], [tex]\( y \to 0 \)[/tex].
### Summary
Here is the result:
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -0.3333 & -1 & -3 & -9 & -27 \\ \hline \end{array} \][/tex]
- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( y \)[/tex]-intercept = [tex]\(-3\)[/tex]
- End Behavior:
- As [tex]\( x \to +\infty \)[/tex], [tex]\( y \to -\infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( y \to 0 \)[/tex]
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