When having to graph a piecewise function, look at each piece separately, making sure to be careful with the restrictions on the variable.

For example,

3x - 2, x less than or equal to -1

f(x) =

-2x + 4, x greater than -1

Graph 3x - 2 first, noting that the largest value that can be used for x is -1. So put -1 in for x to get the endpoint. When doing so, see that f(-1) = (3)(-1) - 2 = -5. Place the coordinate (-1, -5) on the graph. Then use the slope to find another point. Notice on the graph the point found is (-2, -8). Then complete the graph for that portion of the function. Notice the endpoint is a solid circle since it is included in the solution.

Now, graph -2x + 4 the same way. Use -1 to find the endpoint, f(-1) = (-2)(-1) + 4 = 6, so place (-1, 6) on the graph. Find a second point using the slope. Notice on the graph the point found is (0, 4). Then complete the graph for that portion of the function. Notice the endpoint is an open circle since it is not included in the solution.

## Saturday, September 29, 2012

## Thursday, September 27, 2012

A total of 12,000 is invested in two funds paying 9% and 11% simple interest. If the yearly interest is $1,180, how much of the $12,000 is invested at 9% and how much is invested at 11%?

Let's say x dollars are invested at 9% . That means 12,000 - x dollars are invested at 11%. The interest at 9% is x(.09) and the interest at 11% is (12,000 - x)(.11). Add these and set equal to the total amount of interest of 1,180 = .09x + 1320 - 0.11x. 140 = .02x. x= 7000. So 7000 is invested at 9% and 5000 is invested at 11%

How many pounds of M&Ms that cost $7.50 per pound must be mixed with 24 pounds of Reese's Pieces that cost $3.00 per pound to make a mixture that sells for $6.75 per pound?

M&M's 7.50 per pound times x pounds = total cost of 7.50x. The 24 pounds of Reese's Pieces cost a total of 72 dollars. The total mixtures of 6.75 per pound will be 24 + x pounds. So we solve this by takings 7.50x + 72 = 6.75(24 + x). 7.50x + 72 = 162 + 6.75x. 7.50x = 90 + 6.75x. 0.75x = 90, x = 120. You will need 120 pounds of M&M's

## Saturday, September 22, 2012

I was working with a student today who was having problems graphing a system of linear equations. So, I am posting a small part of my book explaining the technique to graph such problems.

Systems of Linear Inequalities

Solving systems of linear inequalities graphically is very similar to solving systems of linear equations. We

graph the system inequalities on the same rectangular coordinate system. Shade the area which represents the intersection of the graphs. To check, we can pick a point from the shaded region and make sure the coordinates satisfy both inequalities in the system. We will start by graphing both separately and then graphing both on the same rectangular coordinate system.

Example: Solve the system by graphing

y ≥ 2x + 3

3x + y < 6

The graph of y ≥ 2x + 3.

Find the shaded region by picking a point on either side of the line to see if it satisfies the inequality. The

easiest to choose is (0, 0). Substitute into y ≥ 2x + 3 to get 0 ≥ 2(0) + 3. This is a false statement so the solution is on the other side of the line. We can check to make sure the shaded region is correct by choosing a point in that region and see if it satisfies the inequality. We'll choose the point (-3, 0). Therefore 0 ≥ 2(-3) + 3, 0 ≥ -3 is a true statement, so our solution is correct.

Next we will graph 3x + y < 6.

If we choose any point in the unshaded region and substitute into both inequalities, we will notice that the

inequalities will not be satisfied. The shaded area in the graph below is the region that only satisfies 3x + y < 6. We find the shaded region by picking a point on either side of the line to see if it satisfies the inequality. Choose (0, 0) since it's the easiest to work with. Therefore 3(0) + 0 < 6 is a true statement and is part of the solution set. All points on the same side of the dotted line satisfy the inequality.

By graphing both inequalities on the same rectangular coordinate system, we will see where the two solutions

intersect.

The large shaded area where the inequality y ≥ 2x + 3 is located satisfies both equations. Therefore it is the

solution to the system of inequalities.

•Note that we show that the green area is the solution by picking a point in the region and testing it in

both inequalities. Any point in that region will satisfy both inequalities

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