We have q < 0 < p < -r
Since q < 0, this means q is some negative number. The ordered pair (x,y) = (2,q) tells us that the point is below the x axis when x = 2. This does not fit graph B because it appears that an x intercept is at x = 2 (though the graph is a bit small so please provide larger versions of each graph if possible).
This does fit graph C because this point is below the x axis when x = 2.
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Then we have 0 < p or p > 0. So p is some positive number. The point (4,p) is some point above the x axis. We started with (2,q) being below the x axis and now we're above the x axis at (4,p). Only graph C fits this description.
Finally p < -r, so -r is some positive number larger than p. Therefore r itself is some negative number. Example: r = -10, -r = -(-10) = 10.
This means the point (6,r) is below the x axis. Once again this fits the description of what graph C is showing.
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<h3>So that is why graph C is the final answer.</h3>
Graphs A and D are ruled out because there is no transition from positive to negative or vice versa. Graph B is ruled out because of the x intercept at 2 (when that point should be below the x axis); however, the graph is a bit small so please provide a larger copy if possible. Thank you.
