Question
Consider the graph of the function h(t)=−6+2/3x
. Which quadrant will the graph not go through and why?(1 point)
Responses
Quadrant III because the slope is negative and the y-intercept is positive.
Quadrant III because the slope is negative and the y-intercept is positive.
Quadrant II because the slope is positive and the y-intercept is negative.
Quadrant II because the slope is positive and the y-intercept is negative.
Quadrant I because the slope is negative and the y-intercept is positive.
Quadrant I because the slope is negative and the y-intercept is positive.
Quadrant IV because the slope is positive and the y-intercept is negative.
. Which quadrant will the graph not go through and why?(1 point)
Responses
Quadrant III because the slope is negative and the y-intercept is positive.
Quadrant III because the slope is negative and the y-intercept is positive.
Quadrant II because the slope is positive and the y-intercept is negative.
Quadrant II because the slope is positive and the y-intercept is negative.
Quadrant I because the slope is negative and the y-intercept is positive.
Quadrant I because the slope is negative and the y-intercept is positive.
Quadrant IV because the slope is positive and the y-intercept is negative.
Answers
GPT-4o mini
To analyze the function \( h(t) = -6 + \frac{2}{3}x \), let's identify its slope and y-intercept.
1. **Identify the slope and y-intercept**:
- The slope \( m = \frac{2}{3} \) is positive.
- The y-intercept is at \( h(0) = -6 \), meaning the graph crosses the y-axis at the point (0, -6).
2. **Determine the quadrants**:
- Because the slope is positive, the function will rise as \( x \) increases.
- The y-intercept at (0, -6) is located in Quadrant IV, where y-values are negative.
3. **Consider the behavior of the graph**:
- For \( h(t) \):
- As \( x \) approaches negative values, \( h(t) \) will decrease further into negative values (Q IV).
- As \( x \) approaches positive values, \( h(t) \) will increase and eventually cross into Quadrant I (where both x and y are positive).
4. **Assess which quadrants the graph will not touch**:
- The graph will pass through Quadrants I and IV since it goes upward (positive slope) from the y-axis.
- The graph will not go through Quadrant II since \( h(t) \) will always be negative below the line (due to the y-value being negative at the y-intercept and gradually increasing into Quadrant I).
Thus, the correct answer is:
**Quadrant II because the slope is positive and the y-intercept is negative.**
1. **Identify the slope and y-intercept**:
- The slope \( m = \frac{2}{3} \) is positive.
- The y-intercept is at \( h(0) = -6 \), meaning the graph crosses the y-axis at the point (0, -6).
2. **Determine the quadrants**:
- Because the slope is positive, the function will rise as \( x \) increases.
- The y-intercept at (0, -6) is located in Quadrant IV, where y-values are negative.
3. **Consider the behavior of the graph**:
- For \( h(t) \):
- As \( x \) approaches negative values, \( h(t) \) will decrease further into negative values (Q IV).
- As \( x \) approaches positive values, \( h(t) \) will increase and eventually cross into Quadrant I (where both x and y are positive).
4. **Assess which quadrants the graph will not touch**:
- The graph will pass through Quadrants I and IV since it goes upward (positive slope) from the y-axis.
- The graph will not go through Quadrant II since \( h(t) \) will always be negative below the line (due to the y-value being negative at the y-intercept and gradually increasing into Quadrant I).
Thus, the correct answer is:
**Quadrant II because the slope is positive and the y-intercept is negative.**