*The only way you can find the solution from the graph is IF you draw a very neat axis system, IF you draw very neat lines, IF the solution happens to be a point with nice neat whole-number coordinates, and IF the lines are not close to being parallel.*

*The only way you can find the solution from the graph is IF you draw a very neat axis system, IF you draw very neat lines, IF the solution happens to be a point with nice neat whole-number coordinates, and IF the lines are not close to being parallel.*On the plus side, since they will be forced to give you nice neat solutions for "solving by graphing" problems, you will be able to get all the right answers as long as you graph very neatly.

The slope is -1, so go down one unit and over one unit and draw a point.

That creates a point at (1, 199), and if you repeat the process starting with that point, you'll get another point at (2, 198).

These are tiny movements on a big line, so draw one more point at the x-intercept to make sure you've got things nicely graphed in the long run.

If Y = 0, then F will be 200, so draw a point at (200, 0).

is usually the easiest) and decide whether these coordinates satisfy the inequality or not.

If they do, shade the half-plane containing that point. Graph each of the inequalities in the system in a similar way.

To graph the second equation, Y = F – 50, use the y-intercept of -50 to draw the first point at (0, -50).

Since the slope is 1, start at (0, -50), and then go up one unit and over one unit. Repeat the process starting from (1, -49) and you'll get a third point at (2, -48).

So one way to graph a line given its equation is to just find two points on it and to draw a straight line through them. To graph a line, it is necesasry to find two points that satisfy .

Then a smooth curve should be drawn through the zeros accounting for multiple roots and making sure the signs match up (i.e. Luckily the quadratic factors as making the roots and . So picking one point less than and plugging it in will determine whether the graph is above or below the -axis for all on the interval Since is positive, the graph is above the -axis. After synthetic division, the polynomial reduces to .

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