case 1

Day 2 Clip 1

Ms Coombs presents the students with a factored form for a polynomial that includes repeated factors:
y = (x – 2)(x – 2)(x – 5)(x – 7).  She does this to illuminate that functions do not always pass through a zero with a change in sign.

 

Day 2 Clip 2

Continuing from the previous clip, Ms. Coombs works with the given function.  Here, she allows the students to hypothesize and refine each other’s ideas for how to modify the question to make the function negative on both sides of x=2.

 

Day 2 Clip 3

Ms Coombs has the students consider that an alternative form for
y = (x – 3)(x + 1)(x + 4) might exist.   To do this, she uses the “graphing calculator” program to expand the factored form.

 

Day 2 - Clip 3

Ms Coombs has the students consider that an alternative form for y = (x – 3)(x + 1)(x + 4) might exist.   To do this, she uses the “graphing calculator” program to expand the factored form. The students realize that the zeros are still present, but not obvious, in the expanded form.  She then challenges the students to reverse their earlier thinking and turn the graph’s zeros into factors.

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