Teach 180: Boolean Algebra, Graphing Style (Day 90)

When one of my students was taking his PreCalculus exam yesterday, he tried to use his graphing calculator to solve a rational inequality.  During the exam, he brought his graphing calculator up to me and asked me why his calculator was showing what it did.  On the left, he has the graph and on the right is what he entered.  Can you tell what is going on?


It took a few seconds for me to understand what it was graphing.  What were those horizontal segments of y = 0 and y = 1?  If you are having trouble figuring it out, look at the graph without the inequality graphed on top of the one produced by my student.  Can you see it now?

When the function is below the x-axis, we know the y-coordinate is less than 0. When the function is above the x-axis, we know the y-coordinate is greater than 0.  So, if the result of (x+2)(x-1)/(x+5) is negative, we have a true statement to the inequality the student entered and a y-value of 1 is plotted.  If the result of (x+2)(x-1)/(x+5) is positive, we have a false statement to the inequality the student entered and a y-value of 0 is plotted.

If we look at the original graph, we can more easily see the solution to the inequality (x+2)(x-1) < 0.      (x+5) It happens when the horizontal pieces are at y=1.  In interval notation, the solution is (-∞, -5) U [-2, 1].  Note that the student still needs to recognize that -5 is not in the domain of the original function. A special shout out and thanks to H.B. for making me think about boolean algebra, graphing style.