Sure, let's examine the given function [tex]\( f(x) = -2 (0.5)^x \)[/tex] step-by-step.
### Step 1: Calculate [tex]\( f(x) \)[/tex] for each given [tex]\( x \)[/tex] value
For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = -2 (0.5)^{-2} \][/tex]
When raising a fraction to a negative power, you take the reciprocal and make the exponent positive:
[tex]\[ (0.5)^{-2} = \left(\frac{1}{0.5}\right)^2 = 2^2 = 4 \][/tex]
So:
[tex]\[ f(-2) = -2 \cdot 4 = -8 \][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -2 (0.5)^{-1} \][/tex]
[tex]\[ (0.5)^{-1} = \frac{1}{0.5} = 2 \][/tex]
So:
[tex]\[ f(-1) = -2 \cdot 2 = -4 \][/tex]
For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -2 (0.5)^0 \][/tex]
Any non-zero number to the power of 0 is 1:
[tex]\[ (0.5)^0 = 1 \][/tex]
So:
[tex]\[ f(0) = -2 \cdot 1 = -2 \][/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -2 (0.5)^1 \][/tex]
[tex]\[ (0.5)^1 = 0.5 \][/tex]
So:
[tex]\[ f(1) = -2 \cdot 0.5 = -1 \][/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = -2 (0.5)^2 \][/tex]
[tex]\[ (0.5)^2 = 0.25 \][/tex]
So:
[tex]\[ f(2) = -2 \cdot 0.25 = -0.5 \][/tex]
Now we can fill in the table:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 \\
\hline
f(x) & -8 & -4 & -2 & -1 & -0.5 \\
\hline
\end{tabular}
\][/tex]
### Step 2: Determine the [tex]\( y \)[/tex]-intercept
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]. We already calculated [tex]\( f(0) \)[/tex] as follows:
[tex]\[ f(0) = -2 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is [tex]\(-2\)[/tex].
### Step 3: Determine the End Behavior
As [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} -2 (0.5)^x \][/tex]
As [tex]\( x \)[/tex] increases, [tex]\( (0.5)^x \)[/tex] becomes very small (approaches 0). Therefore:
[tex]\[ \lim_{x \to \infty} f(x) = -2 \cdot 0 = 0 \][/tex]
Hence:
[tex]\[ y \to 0 \][/tex]
As [tex]\( x \to -\infty \)[/tex]:
[tex]\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} -2 (0.5)^x \][/tex]
As [tex]\( x \)[/tex] becomes more negative, [tex]\( (0.5)^x \)[/tex] grows very large (since [tex]\( 0.5^{-x} \)[/tex] is actually [tex]\( 2^x \)[/tex]). Therefore:
[tex]\[ \lim_{x \to -\infty} f(x) = -\infty \][/tex]
Hence:
[tex]\[ y \to -\infty \][/tex]
### Complete Solution:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 \\
\hline
f(x) & -8 & -4 & -2 & -1 & -0.5 \\
\hline
\end{tabular}
\][/tex]
[tex]\( a = -8, b = -4, \)[/tex] y-intercept [tex]\( = -2 \)[/tex]
End Behavior:
As [tex]\( x \to \infty, y \to 0 \)[/tex]
As [tex]\( x \to -\infty, y \to -\infty \)[/tex]
