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Question
analyzing the graph of a linear function
consider the graph of the linear function ( h(x) = -6 + \frac{2}{3}x ). which quadrant will the graph not go through and why?
- quadrant i, 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 iii, because the slope is negative and the ( y )-intercept is positive
- quadrant iv, because the slope is positive and the ( y )-intercept is negative
Step1: Identify slope and y-intercept
The linear function is $h(x) = -6 + \frac{2}{3}x$, which can be rewritten as $h(x) = \frac{2}{3}x - 6$. Here, slope $m = \frac{2}{3}$ (positive), y-intercept $b = -6$ (negative).
Step2: Analyze quadrant passage
A line with positive slope rises from left to right. A negative y-intercept means it crosses the y-axis below the origin. Starting from $(0, -6)$ (Quadrant IV) and rising right, it goes through Quadrant I; moving left from $(0, -6)$, it goes through Quadrant III. It never enters Quadrant II.
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Quadrant II, because the slope is positive and the y-intercept is negative