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To determine which quadrant the graph of the linear function [tex]\( h(x) = -6 + \frac{2}{3}x \)[/tex] will not go through, let's carefully examine the properties of the line based on the given information. We'll primarily focus on the slope and the y-intercept.
### Understanding the Slope and Y-Intercept
For the linear function [tex]\( h(x) = -6 + \frac{2}{3}x \)[/tex]:
1. Slope (m): The slope [tex]\(\frac{2}{3}\)[/tex] is positive.
2. Y-Intercept (b): The y-intercept is [tex]\(-6\)[/tex], which is negative.
### Effect of Slope and Y-Intercept on Quadrants
A line graph depends on its slope and y-intercept. Here's a brief review of the four quadrants in the coordinate plane:
- Quadrant I (First Quadrant): [tex]\( x > 0 \)[/tex] and [tex]\( y > 0 \)[/tex]
- Quadrant II (Second Quadrant): [tex]\( x < 0 \)[/tex] and [tex]\( y > 0 \)[/tex]
- Quadrant III (Third Quadrant): [tex]\( x < 0 \)[/tex] and [tex]\( y < 0 \)[/tex]
- Quadrant IV (Fourth Quadrant): [tex]\( x > 0 \)[/tex] and [tex]\( y < 0 \)[/tex]
Now let's analyze the behavior of the line based on its slope and y-intercept:
1. Positive Slope: This indicates that as [tex]\( x \)[/tex] increases, [tex]\( y \)[/tex] also increases. It results in a line that moves from the bottom left to the top right (i.e., rising line).
2. Negative Y-Intercept: This means the line crosses the y-axis below the origin.
With a positive slope and a negative y-intercept:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = -6 \)[/tex], so the line starts below the origin on the y-axis.
- As [tex]\( x \)[/tex] increases from 0, [tex]\( y \)[/tex] will increase (due to the positive slope), causing the line to rise into Quadrant I.
- As [tex]\( x \)[/tex] decreases from 0, [tex]\( y \)[/tex] will also decrease (since the line intercepts below the origin), causing it to drop further into Quadrant III.
### Examination of Quadrants
1. Quadrant I: The line goes through this quadrant because as [tex]\( x \)[/tex] increases to the right of the y-axis, [tex]\( y \)[/tex] becomes positive.
2. Quadrant III: The line goes through this quadrant because as [tex]\( x \)[/tex] decreases to the left of the y-axis, [tex]\( y \)[/tex] becomes negative, continuing the drop from the negative y-intercept.
3. Quadrant IV: The line also intersects this quadrant as moving to the right of the y-axis ([tex]\( x \)[/tex] positive), while [tex]\( y = -6 + (positive value) \)[/tex] is still negative, and then increases up through the x-axis into Quadrant I.
4. Quadrant II: The line will not go through this quadrant because:
- For this line to enter Quadrant II, it would need to rise from the positive y-axis location with a negative x-intercept movement. However, our line starts below the y-axis (negative intercept) and moves upwards to Quadrant I and downwards to Quadrant III.
- The positive slope always ensures that while increasing [tex]\( x \)[/tex] (going right), [tex]\( y \)[/tex] will eventually turn positive, bypassing Quadrant II entirely since it does not rise leftwards (negative x) from a positive y-direction.
### Conclusion
The graph of [tex]\( h(x) = -6 + \frac{2}{3}x \)[/tex] will not go through Quadrant II, because the slope is positive and the y-intercept is negative. Therefore, the correct answer is:
Quadrant II, because the slope is positive and the y-intercept is negative.
### Understanding the Slope and Y-Intercept
For the linear function [tex]\( h(x) = -6 + \frac{2}{3}x \)[/tex]:
1. Slope (m): The slope [tex]\(\frac{2}{3}\)[/tex] is positive.
2. Y-Intercept (b): The y-intercept is [tex]\(-6\)[/tex], which is negative.
### Effect of Slope and Y-Intercept on Quadrants
A line graph depends on its slope and y-intercept. Here's a brief review of the four quadrants in the coordinate plane:
- Quadrant I (First Quadrant): [tex]\( x > 0 \)[/tex] and [tex]\( y > 0 \)[/tex]
- Quadrant II (Second Quadrant): [tex]\( x < 0 \)[/tex] and [tex]\( y > 0 \)[/tex]
- Quadrant III (Third Quadrant): [tex]\( x < 0 \)[/tex] and [tex]\( y < 0 \)[/tex]
- Quadrant IV (Fourth Quadrant): [tex]\( x > 0 \)[/tex] and [tex]\( y < 0 \)[/tex]
Now let's analyze the behavior of the line based on its slope and y-intercept:
1. Positive Slope: This indicates that as [tex]\( x \)[/tex] increases, [tex]\( y \)[/tex] also increases. It results in a line that moves from the bottom left to the top right (i.e., rising line).
2. Negative Y-Intercept: This means the line crosses the y-axis below the origin.
With a positive slope and a negative y-intercept:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = -6 \)[/tex], so the line starts below the origin on the y-axis.
- As [tex]\( x \)[/tex] increases from 0, [tex]\( y \)[/tex] will increase (due to the positive slope), causing the line to rise into Quadrant I.
- As [tex]\( x \)[/tex] decreases from 0, [tex]\( y \)[/tex] will also decrease (since the line intercepts below the origin), causing it to drop further into Quadrant III.
### Examination of Quadrants
1. Quadrant I: The line goes through this quadrant because as [tex]\( x \)[/tex] increases to the right of the y-axis, [tex]\( y \)[/tex] becomes positive.
2. Quadrant III: The line goes through this quadrant because as [tex]\( x \)[/tex] decreases to the left of the y-axis, [tex]\( y \)[/tex] becomes negative, continuing the drop from the negative y-intercept.
3. Quadrant IV: The line also intersects this quadrant as moving to the right of the y-axis ([tex]\( x \)[/tex] positive), while [tex]\( y = -6 + (positive value) \)[/tex] is still negative, and then increases up through the x-axis into Quadrant I.
4. Quadrant II: The line will not go through this quadrant because:
- For this line to enter Quadrant II, it would need to rise from the positive y-axis location with a negative x-intercept movement. However, our line starts below the y-axis (negative intercept) and moves upwards to Quadrant I and downwards to Quadrant III.
- The positive slope always ensures that while increasing [tex]\( x \)[/tex] (going right), [tex]\( y \)[/tex] will eventually turn positive, bypassing Quadrant II entirely since it does not rise leftwards (negative x) from a positive y-direction.
### Conclusion
The graph of [tex]\( h(x) = -6 + \frac{2}{3}x \)[/tex] will not go through Quadrant II, because the slope is positive and the y-intercept is negative. Therefore, the correct answer is:
Quadrant II, because the slope is positive and the y-intercept is negative.
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