Question 1124833
<br>
|x| is always 0 or positive; so
3|x| is always 0 or positive; so
3|x|+1 is always positive<br>
So the graph of the EQUATION y = 3|x|+1 is always above the x-axis, which means it is never in quadrants III or IV.<br>
And then since the graph of y = 3|x|+1 is never in quadrants III or IV, the graph of the inequality y > 3|x|+1 is also never in quadrants III or IV.<br>
Finally, there are no restrictions on the value of x, so x can be either positive or negative (or 0); so the graphs of the equation AND the inequality are in both quadrants I and II.<br>
Answer B.