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Consider the graph of the linear function [tex]h(x)=-6+\frac{2}{3} x[/tex]. Which quadrant will the graph not go through and why?

A. Quadrant I, because the slope is negative and the [tex]y[/tex]-intercept is positive
B. Quadrant II, because the slope is positive and the [tex]y[/tex]-intercept is negative
C. Quadrant III, because the slope is negative and the [tex]y[/tex]-intercept is positive
D. Quadrant IV, because the slope is positive and the [tex]y[/tex]-intercept is negative


Sagot :

To determine which quadrant the graph of the linear function [tex]\( h(x) = -6 + \frac{2}{3} x \)[/tex] will not go through, we follow these steps:

1. Identify the slope and y-intercept:
The given linear function is in the form [tex]\( h(x) = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
[tex]\[ m = \frac{2}{3}, \][/tex]
[tex]\[ b = -6. \][/tex]

2. Understand the y-intercept:
The y-intercept [tex]\( b \)[/tex] is the point where the graph crosses the y-axis. Here, it is [tex]\( -6 \)[/tex], which is below the origin on the y-axis.

3. Interpret the slope:
The slope [tex]\( \frac{2}{3} \)[/tex] is positive, meaning the line rises as it moves from left to right.

4. Determine which quadrants the line will pass through based on the slope and y-intercept:
- Since the y-intercept is negative ([tex]\( -6 \)[/tex]), the graph starts below the origin on the y-axis.
- As the slope is positive ([tex]\( \frac{2}{3} \)[/tex]), the line will rise as it moves rightward.

Therefore:
- Starting in the negative y-region (below the x-axis), the graph will initially be in Quadrant III when moving from left to right.
- Because the slope is positive, the line will move up and to the right but still remain below the x-axis, entering Quadrant IV.

5. Which quadrants does the graph never touch?:
- The graph rises but starts below the x-axis and crosses the x-axis to enter Quadrant IV directly.
- It never crosses into the quadrants where [tex]\( y \)[/tex] is positive (Quadrants I and II) since the y-intercept is negative and the slope is insufficient to bring [tex]\( y \)[/tex] positive.

With this understanding, the correct answer to which quadrant the graph will not pass through given its positive slope and negative y-intercept is:

Quadrant I, because the slope is positive and the y-intercept is negative.