Question

complete the table for the function $y = \frac{1}{3}(9)^x$.

x	y
-1
0
1
2

now, graph the function.
plot two points to graph the function.
(graph with x-axis from -9 to 9 and y-axis from -27 to 27, grid lines.)

Explanation:

Step1: Calculate y when x = -1

Substitute \( x = -1 \) into \( y = \frac{1}{3}(9)^{x} \). We know that \( 9^{-1} = \frac{1}{9} \), so \( y = \frac{1}{3} \times \frac{1}{9} = \frac{1}{27} \)? Wait, no, wait. Wait, \( 9^{-1} \) is \( \frac{1}{9} \), but \( \frac{1}{3} \times \frac{1}{9} = \frac{1}{27} \)? Wait, no, maybe I made a mistake. Wait, \( 9^{x} \) when \( x = -1 \) is \( 9^{-1} = \frac{1}{9} \), then multiply by \( \frac{1}{3} \), so \( \frac{1}{3} \times \frac{1}{9} = \frac{1}{27} \)? Wait, no, that seems small. Wait, maybe I messed up the exponent. Wait, \( 9^{x} \) is \( 9 \) to the power of \( x \). Let's recheck.

Wait, \( x = -1 \): \( y = \frac{1}{3}(9)^{-1} = \frac{1}{3} \times \frac{1}{9} = \frac{1}{27} \). Hmm.

Step2: Calculate y when x = 0

Substitute \( x = 0 \) into the function. Any number to the power of 0 is 1, so \( 9^{0} = 1 \). Then \( y = \frac{1}{3} \times 1 = \frac{1}{3} \)? Wait, no, wait, \( \frac{1}{3}(9)^0 = \frac{1}{3} \times 1 = \frac{1}{3} \)? Wait, that seems correct. Wait, but maybe I made a mistake here. Wait, \( 9^0 = 1 \), so \( \frac{1}{3} \times 1 = \frac{1}{3} \).

Step3: Calculate y when x = 1

Substitute \( x = 1 \): \( 9^1 = 9 \), so \( y = \frac{1}{3} \times 9 = 3 \).

Step4: Calculate y when x = 2

Substitute \( x = 2 \): \( 9^2 = 81 \), so \( y = \frac{1}{3} \times 81 = 27 \).

Wait, let's verify each step again:

• For \( x = -1 \): \( 9^{-1} = \frac{1}{9} \), so \( \frac{1}{3} \times \frac{1}{9} = \frac{1}{27} \). Correct.
• For \( x = 0 \): \( 9^0 = 1 \), so \( \frac{1}{3} \times 1 = \frac{1}{3} \). Correct.
• For \( x = 1 \): \( 9^1 = 9 \), \( \frac{1}{3} \times 9 = 3 \). Correct.
• For \( x = 2 \): \( 9^2 = 81 \), \( \frac{1}{3} \times 81 = 27 \). Correct.

So the table values are:

• \( x = -1 \): \( y = \frac{1}{27} \)
• \( x = 0 \): \( y = \frac{1}{3} \)
• \( x = 1 \): \( y = 3 \)
• \( x = 2 \): \( y = 27 \)

Now, for graphing, we can plot two points. Let's choose \( (0, \frac{1}{3}) \) and \( (1, 3) \), or \( (1, 3) \) and \( (2, 27) \), but \( (2, 27) \) is high. Alternatively, \( (0, \frac{1}{3}) \) and \( (-1, \frac{1}{27}) \), but those are close to the x-axis. Maybe better to use \( (0, \frac{1}{3}) \) and \( (1, 3) \).

Answer:

The table values are:

• When \( x = -1 \), \( y = \frac{1}{27} \)
• When \( x = 0 \), \( y = \frac{1}{3} \)
• When \( x = 1 \), \( y = 3 \)
• When \( x = 2 \), \( y = 27 \)

For graphing, two points can be \( (0, \frac{1}{3}) \) and \( (1, 3) \) (or other pairs from the table).