Chapter 5: Data analysis and findings
5.3 Does the use of dynamic graphing software such as Geogebra enhance conceptual
5.3.3 Results and findings for question 6
Researcher: Are you happy with your final sketch?
Gerald: Yes, sir.
Researcher: Ok please explain to me how you arrived at it.
Gerald: Well, because the gradient is positive (pointing to the segment of the graph in the third quadrant) it has to be above the x-axis and also above the turning point.
Researcher: Are you saying that the gradient of this graph is always going to be positive?
Gerald: Yes.
Researcher: Ok, look at the graph are there any points along it where the tangent will have a negative slope?
Gerald: Over here it will be negative (correctly identifying the segment)
Researcher: Ok, that is correct but have you accounted for that segment in your sketch?
Gerald: No, Sir.
Researcher: Would you like to reconsider your sketch then?
Gerald: I am very confused now, Sir.
Gerald’s approach relied on the degree of the polynomial. Although he finally deduced that the resulting gradient graph was concave up, he failed to incorporate other relevant information. For instance, he did acknowledge that the graph had a segment with a negative gradient but this was ignored when drawing the final sketch. This tendency to ignore relevant information because it poses a problem when it comes to sketching a graph was also observed by Baker et al. (2000).
Taken from Park (2013, p.639) In the main, two strategies were observed as the pupils attempted to solve the problem. In total four students managed to make the correct choice. Two students, Allen and Gerald, first deduced that the given graph represented a quadratic function hence the required graph from the given distracters had to be cubic. They then used their knowledge of cubic graphs to make the correct choice. The other two students, Adam and Wayne, took it a step further and attempted to explain their choice of graph and made explicit reference to the gradient of the chosen function graph at different points. Gestures and traces made often mimicked the orientation of the tangent along the graph similar to what they observed during the dragging exercise. Arguably their success could be attributed to their experience with Geogebra experience. Mark did not engage with the problem while, Tinashe tried but eventually gave up after failing to coordinate the pertinent information required to make a choice.
The case of Gerald
Gerald eliminated (a) and (b) by saying that the required function had to have two turning points. These graphs, he further argued, were graphs of linear functions.
Gerald: The parabola that we are given represents a sad face (concave down) and so the required graph must have a negative gradient.
Researcher: What do you mean a negative gradient?
Gerald: I mean a graph with a negative x³ and I think it is (c) because of the shape of the graph. It starts from the second quadrant going into the fourth quadrant and that’s the graph of a negative x³.
The case of Adam
Adam: We know that its (given graph) a parabola with a negative a value.
Researcher: Ok.
Adam: We know that it (required answer) has to be an x³ graph so we can cancel out (a) and (b). We now look for the x intercepts on the parabola those will be the places on the cubic function where we have a zero gradient.
Researcher: Yes?
Adam: I am going to choose (c) because if you look at these points (turning points) they are at negative one and two and the gradient at these points is zero. That means the matching point (on gradient function) y has to be zero.
Just to further check his understanding,
Researcher: Ok, but why are you discarding this one (referring to d)? It also has a turning point.
Adam: Yes but not at 2, its gradient is not zero at two.
Adam’s reasoning was consistent with what was observed as he solved question 5b. He considered not only the degree of the polynomial but also the value of the gradient along the chosen graph and compared it successfully with the given parabola.
The case of Wayne
Wayne: I am going to cross out (a) and (b) because they represent straight lines with x to the power of one and the corresponding gradient functions would have x to the power of zero. The given graph is a parabola so it is an x², and one of these to be the actual graph it has to be an x³ function.
Researcher: Ok.
Wayne: I will go with (c) based on my earlier argument that if the gradient is negative it has to be below the x-axis, now both (d) and (e) have positive gradients (pointing to segments in the third quadrant) and they should be above the x- axis….(c) has a negative gradient and the parabola also starts below the x-axis.
So I will go with (c).
He went on to match the segment of the graph that had a positive gradient and said that on the gradient function that segment had to be above the x-axis.
Wayne: This part has a positive gradient and it basically corresponds to that entire part of the parabola (tracing out the part of the parabola above the x-axis)
5.4 Findings based on Section 5.3: Does the use of dynamic graphing software such as