To solve this problem, we need to understand how to adjust the linear function to move its graph vertically. The linear function given is [tex]\( h(x) = -\frac{2}{3}x + 5 \)[/tex], which is in the slope-intercept form [tex]\( y = mx + b \)[/tex].
The components of the linear equation are:
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{2}{3} \)[/tex].
- The [tex]\( y \)[/tex]-intercept ([tex]\( b \)[/tex]), the point where the line crosses the [tex]\( y \)[/tex]-axis, is 5.
To move the graph of the function down by 3 units, we need to adjust the [tex]\( y \)[/tex]-intercept ([tex]\( b \)[/tex]) accordingly.
### Step-by-Step Solution:
1. Identify the current [tex]\( y \)[/tex]-intercept:
The current [tex]\( y \)[/tex]-intercept ([tex]\( b \)[/tex]) of the given function [tex]\( h(x) = -\frac{2}{3}x + 5 \)[/tex] is 5.
2. Calculate the new [tex]\( y \)[/tex]-intercept:
To move the graph down by 3 units, we need to subtract 3 from the current [tex]\( y \)[/tex]-intercept.
[tex]\[
\text{New } b = 5 - 3
\][/tex]
[tex]\[
\text{New } b = 2
\][/tex]
3. Construct the new equation:
The new function with the [tex]\( y \)[/tex]-intercept adjusted is:
[tex]\[
h(x) = -\frac{2}{3}x + 2
\][/tex]
Among the given choices:
- Changing [tex]\( x \)[/tex] to -3 does not affect the vertical position of the graph.
- Changing the value of [tex]\( m \)[/tex] to -3 modifies the slope of the line but not its vertical position.
- Changing the value of [tex]\( b \)[/tex] to 2 adjusts the [tex]\( y \)[/tex]-intercept to 2, which moves the graph down by 3 units.
- Changing the value of [tex]\( m \)[/tex] to 2 once again alters the slope, not the vertical position.
### Conclusion:
The correct action to move the graph of the function [tex]\( h(x) = -\frac{2}{3}x + 5 \)[/tex] down by 3 units is to change the value of [tex]\( b \)[/tex] to 2.
Thus, the correct answer is:
The value of [tex]\( b \)[/tex] to 2.
