Question

graph this function:
$y = \frac{1}{2}(3)^x$
pick two points to graph the function.

Explanation:

Step 1: Find the y - intercept (x = 0)

Substitute \( x = 0 \) into the function \( y=\frac{1}{2}(3)^{x} \). Recall that any non - zero number to the power of 0 is 1, so \( 3^{0}=1 \). Then \( y=\frac{1}{2}\times1=\frac{1}{2} \). This gives us the point \( (0,\frac{1}{2}) \).

Step 2: Find another point (x = 1)

Substitute \( x = 1 \) into the function. We know that \( 3^{1}=3 \), so \( y=\frac{1}{2}\times3=\frac{3}{2} \). This gives us the point \( (1,\frac{3}{2}) \).
After finding these two points, we can plot them on the coordinate plane and then sketch the exponential function. Since the base of the exponential function \( 3>1 \), the function is an increasing exponential function, and the coefficient \( \frac{1}{2} \) vertically scales the standard exponential function \( y = 3^{x} \) by a factor of \( \frac{1}{2} \).

Answer:

To graph \( y=\frac{1}{2}(3)^{x} \), we find two points:

• When \( x = 0 \): \( y=\frac{1}{2}(3)^{0}=\frac{1}{2}(1)=\frac{1}{2} \), so the point is \( (0, \frac{1}{2}) \).
• When \( x = 1 \): \( y=\frac{1}{2}(3)^{1}=\frac{3}{2} \), so the point is \( (1, \frac{3}{2}) \).

Plot these points and draw the exponential curve (increasing, since the base \( 3>1 \), scaled by \( \frac{1}{2} \)).