Point-slope Formula and Point-point Formula:

Logic of the Lesson

Pat Thompson

 

The following example is of a lesson logic for teaching the point-slope formula (it really should be called the point-rate formula) and the point-point formula in Algebra I.

 

A lesson logic is an outline of how you will develop the lesson's main ideas. It does not pay attention to time, meaning that the "lesson" may transcend several class periods. It does not give the level of detail that a lesson plan gives, meaning it might not say how you will organize the classroom, how you will transition from one activity to another, etc. Instead, it focuses on the ideas you will develop, the way you develop them, and why you take the approach you take. All lesson logics assume that the lesson unfolds by way of the teacher leading a conceptual conversation around the lesson's central goals.

 

The following lesson logic unfolds these ideas:

 

1.     You can determine the coordinates of all the points on a linear function's graph by knowing just its rate of change and one point through which the graph passes. Thus, you can determine the coordinates of the y-intercept just by knowing the function's rate of change and one point its graph passes through. Therefore, you can put the function in standard form y = mx + b, where m is the function's rate of change and b is its y-intercept.

2.     If you know the coordinates of two points that a linear function's graph passes through, then you can calculate the function's rate of change from those coordinates. Then you are back to the situation in (1), where you know the function's rate of change and the coordinates of one point that its graph passes through.

Meanings students must have before the lesson:

 

1)    The rate of change of a function describes how y changes with respect to changes in  x (ex: if the rate of change is 3.5, when x changes by 1, y changes by 3.5 * 1; when x increases by 1/100, y increases by 3.5 × 1/100);

2)    Lines are collections of an infinite number of points.

 

Step

Action

Reason

1.

Give the students a point in the xy-plane and a rate of change. [Note: Here it may be helpful to give them this rate of change in a context they are familiar with. If they have encountered rates of change when talking about distance a car travels over some amount of time, refer back to such an example so students can relate this to their prior knowledge about the subject of rate of change.] Now, ask the students to find a second point where the rate of change between these two points is the given rate of change.

This forces students to think about the relationship between the x and y coordinates for points on a linear function’s graph. They need to think “If I am given a rate of change, this means that when x increases by some amount, y increases by the rate of change times the amount by which x changes.”

2.

Ask different students in the class the point that they chose and have them explain how they found it. Most likely, different students will find different points that satisfy the above conditions.

This emphasizes the fact that many different pairs of points can share the same rate of change between them.

3.

Assume that at least one of the students found his or her point by finding the value of y when x increased by 1.Ask students what would happen if x changed by 1/100, 2/100, .0375, or by 0.5281? By 7, 90.3, or by 204.8? By -2, -3.4, or by -7.3? Does their reasoning change even though the numbers are different?

This helps students to see once again that y changes by the rates times the amount by which x changes.

4.

Ask the students if they have found all the points that satisfy the above scenario that the rate of change between their new point and the initial point is the given rate of change. If not, ask them where the other points are that do satisfy the scenario.

 

Point out to students that they have established that all the points on a linear function’s graph are determined by knowing just the function’s rate of change and one point on the graph.

Students should see that any point on the line determined by their two points would also satisfy this relationship: However much the x-coordinate of the second point changes from the x-coordinate of the first point, the y-coordinate of the second point changes by the rate times the change in the x-coordinates. Symbolize this as “if x changes by ∆x, then y changes by r∆x.”

5.

Tell them that you are thinking of a function that has a line as its graph. The graph passes through the point (7.5, 2.4), and the function has a rate of change of 3. Ask them, “What will be the value of y when x is 0? Do this again with a different point and a different slope. Focus on the similarity in reasoning between the two cases.

 

This is really to find the y-intercept. Students should eventually reason that moving from (7.5,2.4) is like changing x by -7.5. If the function’s rate of change is 3, then y will change by 3ę(-7.5), or -22.5. So y will change from 2.4 to (2.4-22.5), hence y will be -20.1 when x is 0. Asking students to do it twice will highlight the steps in reasoning over the steps in arithmetic.

 

6.

Ask students, “A function has a line as its graph. The graph passes through the point (a,b) with slope 2.5. What is the value of y when x is 0?

This asks students to generalize to arbitrary points.

7.

Tell students, “The students in another school do not know how to find the y intercept for a linear function when all they’re given is a point on the graph and the function’s rate of change. You do not know the precise points they will work with. Write them a letter to tell them how to do this so that your plan will work for whatever points they pick.”

This step forces students to formalize their method for finding a linear function’s y-intercept. They will need to decide how to represent the given information and how to help the other students to apply this method.

Students should end up with something like:

. If we move from x1 to 0, then y1 will change by r(-x1).

8.

Say, “The graph of a linear function passes through the point (-3,-4.1) with slope r. Write a formula that represents this function in standard form (i.e., in the form y = mx + b).”

This is the point-slope formula. Students should be able to answer this question now that they know how, in general, to calculate the function’s y-intercept.

9.

Say, “The graph of a linear function passes through the point (x1,y1) with slope r. Write a formula that represents this function in standard form (i.e., in the form y = mx + b).”

Ditto the above, except this is the most general case.

10.

Tell students, “I’m thinking of a linear function whose graph passes through the points (7.2, 5.5) and (12.7, 2.1). Write a formula that represents this function in standard form (i.e., in the form y = mx + b).”

They have only the added step of determining the function’s rate of change. Once they have this, they also have one point the graph passes through, so the problem is reduced to the prior case.

11.

Tell students, “I’m thinking of a linear function whose graph passes through the points (x1,y1) and (x2,y2). Write a formula that represents this function in standard form (i.e., in the form y = mx + b).”

This is an occasion to generalize from the concrete example in 10. The function's rate of change r is
(
y2 y1)/( x2x2) and it passes through the point (x2,y2) (or through the point (x1,y1)).

12.

Recap what they have done:

  1. Established that for a linear function, being given one point and the function’s rate of change is sufficient to determine all the points on the function’s graph.
  2. Developed a method by which to find a linear function’s y-intercept when given the function’s rate of change and one point that its graph passes through.
  3. Used that method to develop a formula for writing a linear function in point-slope form when all they know is the function’s rate of change and a point its graph passes through.
  4. Used their reasoning in (2 and 3) to develop a formula for writing a linear function in point-slope form when all they know is two points that the functions graph passes through.