Question 709377
Pretend for a moment that the function was simply g(x) = -x then I hope you would recognize this as the equation for a line. A line in slope-intercept form. A line with a slope of -1 and a y-intercept of 0. One of the points on this pretend line is (3, -3)<br>
Since the real g(x) is -x for all x's except 3, graph the line described above <i>except for the point (3, -3)!</i>. In other words. draw an small open circle (a small ring) at the point (3, -3). (This open circle indicates that this point is <i>not</i> included.) Then draw the rest of the line as normal. When you are finished it should look like the the line y = -x except that it has a "hole" in it at (3, -3).<br>
To finish the graph just plot the point (3, 2). (This is a regular point, a dot, not the open circle we used at (3, -3).)<br>
Your final graph should be a line with a hole in it at (3, -3) plus the extra point (3, 2) (which is nowhere near the line).