Tag Archives: systems of equations

180 Blog Algebra 2, Blog

Confirming Answers with Graphing Software

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I added two components to a lesson task on rational equations. The first was an idea called Hint Cards which I discussed in a previous post. The second one was having students use Desmos to confirm numerical answers that they found algebraically. This lesson is worth sharing, because of what I discovered that the students didn’t know how to do and what they learned from the activity.

The students had just completed the  Optimum Bait Company task. I gave them 5 minutes to check their answers with Desmos. Mathematically speaking, they needed to confirm that for the equation, , the following were true:

  • C(1000) = 4.45
  • C(4000) = 1.3, C(8000) = 0.775, C(12000) = 0.6 and C(42000) = 0.26
  • The horizontal asymptote is y = 0.25
  • C(5600) = 1

I was struck by how utterly stumped they were on using a graph to check these answers, even though they possessed all the prerequisite knowledge necessary. They understood the context of their answers (The average production cost of 1000 lures is $4.45). They knew how to find a y-value from a graph given the x-value (C(1000) = 4.45). They also were experienced at graphing equations with Desmos (y = (4200 + 0.25x)/x), and they knew how to establish a domain and range for axis in context. While my students had each of these four connected skills, they were missing the connection between them all, so I embraced the teachable moment.

I started with the basic idea that the equation we need to graph was the one for the average cost:  C(x) = (4200 + 0.25x)/x. A major problem faced them, though, when they first graphed this equation… they couldn’t see the rational function in the default window.

Bait Desmos Blank

So my next move with the class, was to use the numeric results to determine the domain and range. The range was simple since we all of the monetary answers were between 0 and $5.00. The domain needed a bit more discussion because the one value of 42000 lures compressed the graph so significantly that the class thought it better to leave it out of the visible domain, so we agreed upon 0 < x < 13000.

Bait Desmos Window

From here I could have just traced with finger along the screen to show where a point with a value of x would be located on the curve, but I wanted to tie in the writing of horizontal and vertical equations, and solutions of systems. Therefore, I had the students enter the additional equation of x = 1000, and click on the point of intersection.

Bait Desmos 1K

The students were getting happier and more confident so we kept rolling by entering a table with the additional increasing values of x, representing the number of lures.

Bait Desmos Table

The table confirmed the students answers and supported their common response that the values were approaching a limit of 25 cents. This appealed to their sense of context that the cost per lure could not drop below the original 25 cents per lure. That made for an easy connection to the finding the horizontal asymptote of the rational function for which the degree of the numerator and denominator are equal, which in this case was also 0.25. So we graphed this as an asymptote. It turns out that it was easier to see how closely the curve approached the asymptote if we temporarily increased our visible domain to the 42000 lures.

Bait Desmos Asymptote

Finally, we entered the equation y = 1 in order to see the number of lures required to drive the average cost down to $1 a lure.

Bait Desmos 1 Dollar

This serendipitous exercise was amazingly productive for reinforcing understanding of graphing in general, as well as the connection between numerical values, algebraic formulas and context.

Blog, Lesson: Algebra 2

Hint Cards

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Hint CardI added two components to a lesson task on rational equations.The first was an idea called, Hint Cards, shared by Michael Pershan at Shadow Con at NCTM 2015.  The second, which I will discuss in another post, was having students use Desmos to confirm answers that they found algebraically. There were so many surprising positives from this lesson that I have to share.

As always, lesson awesomeness starts with a good task. The class had studied the simplifying, solving and graphing of rational functions. It was time to write and apply them. My school’s Algebra 2 team decided on a common task created by our colleague, Jake Paino, titled Optimum Bait Company. The task offered the following context followed by six prompts.

My brother Matt owns Optimum Bait Company. Optimum Bait Company manufactures fishing lures. The monthly cost to run the factory is $4200 and the cost of producing each lure is an additional $0.25 per lure.

  1. If he produces 1000 lures in one month, what is the average production cost per lure?
  2. Create a function, C(x), that models the average production cost per lure.
  3. Calculate the average production cost per lure if he produces 4000 lures in one month? 8000 lures? 12000 lures? 420000 lures?
  4. As he produces more lures what price does the average cost of production approach? Why?
  5. If he wants the average cost of production to be $1, how many lures would he have to produce in one month?
  6. If he wants to make a profit of at least $4000 per month, what is the minimum number of lures he would have to produce if he sells every lure he produces for $4?

I was thinking that the students would need a lot of help on this, so I created a set of six Hint Cards. Each card gave some assistance for one of the prompts.

Front of Card

 Back of Card

#1: Average Cost of 100 lures

 Average = Total Cost/Total Number

#2: Create C(x)

 Let x = number of lures

#3: Average Cost per Lure

C(4000) = (4200 + 0.25(4000))/4000
#4: Limit of Average Cost

The Ratio of the Leading Coefficients

#5: Average Cost of $1

C(x) = 1, instead of   x = 1
#6: Profit of $4000

Profit = Income – Expenses

As an incentive, I announced the following scoring system.

  • Like all other tasks, this will be worth 5 points.
  • There are 6 prompts. Every wrong answer to a prompt costs a point.
  • There are also 6 hints. Every hint used costs a point.
  • Yes, that means you either have one free pass on a wrong answer, or a free hint.
  • The only thing that you may ask of the teacher is for a hint card to a specific prompt.
  • 30 minutes will be allotted to complete the task.

I was pleasantly surprised how little they needed or wanted help. Many groups didn’t even take advantage of their free hint. In fact, for all of eight groups, I gave out only seven Hint Cards… total. The most common hints asked for were for #1 and #5. The most commonly missed questions were the last two. I suspect that if I had given more time on the task, students may have ask for more hints and given more effort on what I consider to be the hardest questions for the task. That’s a lesson for next time; there will definitely be a next time because of the benefits that resulted from the Hint Cards technique:

  • The time crunch spurred a hyper-focus in the students.
  • The level and intensity of the student discourse was heightened tremendously.
  • A common dynamic was having one student raise a hand for a hint, while another group member protested, “we don’t need it, yet.”
  • The average score on the task was 4.2 out of 5.

Hint Cards delivered a terrific learning experience for my students. One that I will be sure to give them again.