To complete the table for the function \( y = 8\left(\frac{1}{4}\right)^x \), we can calculate the values of \( y \) for the given values of \( x \).
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For \( x = -2 \): \[ y = 8\left(\frac{1}{4}\right)^{-2} = 8 \times 4^2 = 8 \times 16 = 128 \]
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For \( x = -1 \): \[ y = 8\left(\frac{1}{4}\right)^{-1} = 8 \times 4^1 = 8 \times 4 = 32 \]
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For \( x = 0 \): \[ y = 8\left(\frac{1}{4}\right)^{0} = 8 \times 1 = 8 \]
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For \( x = 1 \): \[ y = 8\left(\frac{1}{4}\right)^{1} = 8 \times \frac{1}{4} = 2 \]
Now let's fill out the table:
| x | y | | --- | --- | | -2 | 128 | | -1 | 32 | | 0 | 8 | | 1 | 2 |
Points to graph
Now let's take two points to graph the function. We can take:
- \( (-2, 128) \)
- \( (1, 2) \)
Graphing
- The point \( (-2, 128) \) will be very high on the y-axis.
- The point \( (1, 2) \) is much lower and easy to plot.
To visualize this, you would plot the two points on a coordinate system, then note that the function will continually decrease as \( x \) increases, due to the nature of the function \( y = 8\left(\frac{1}{4}\right)^x \), which is an exponential decay function.
If you were to graph it using a graphing tool, you would find that the curve approaches the x-axis but never touches it, representing that as \( x \to \infty \), \( y \to 0 \).