I am working with 8th graders on quadratic equations. When I give them a table of values, we can determine if the rule is quadratic by looking for a constant difference between y-value differences, ie; the 2nd difference is constant. We then use this to find the "a" value by taking the 2nd difference and dividing it by 2. Then we use the y intercept and input/output values to write the rule in "y=ax^2 + bx + c" form. My question is: how can I explain to them why taking the 2nd difference and dividing by 2 gives the "a" value? A simple qualitative description (without getting into differential equations?) is what I'm looking for; they are 1st year algebra students."
Quadratic Equations
- Thread starter chouse
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I forgot to mention - my question above is related to another problem I am having: I want to explain to them why the falling object formula is h = -16t^2 +vt + s.
They know that acceleration due to gravity is -32 ft/s/s, and we can see that the "a" value is half of -32. So of course they want to know why the formula is not h = -32t^2 + vt + s.
They know that acceleration due to gravity is -32 ft/s/s, and we can see that the "a" value is half of -32. So of course they want to know why the formula is not h = -32t^2 + vt + s.
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This could probably be related to calculus and derivatives, but that might be too advanced for your audience.
You could always show the algebra. For simplicity, use x - 1, x, and x + 1 for your three "typical" x-values. Then your three quadratic terms are:
. . . . .\(\displaystyle \mbox{first term: }\, a(x\, -\, 1)^2\, +\, b(x\, -\, 1)\, +\, c\, =\, ax^2\, -\, 2ax\, +\, bx\, +\, a\, -\, b\, +\, c\)
. . . . .\(\displaystyle \mbox{second term: }\, ax^2\, +\, bx\, +\, c\)
. . . . .\(\displaystyle \mbox{third term: }\, a(x\, +\, 1)^2\, +\, b(x\, +\, 1)\, +\, c\, =\, ax^2\, +\, 2ax\, +\, bx\, +\, a\, +\, b\, +\, c\)
Then subtract the first from the second:
. . . . .\(\displaystyle \left(ax^2\, +\, bx\, +\, c \right)\, -\, \left(ax^2\, -\, 2ax\, +\, bx\, +\, a\, -\, b\, +\, c \right)\)
. . . . .\(\displaystyle =\, ax^2\, -\, ax^2\, +\, bx\, -\, bx\, +\, c\, -\, c\, +\, 2ax\, -\, a\, +\, b\)
. . . . .\(\displaystyle =\, 2ax\, -\, a\, +\, b\)
...and the second from the third:
. . . . .\(\displaystyle \left(ax^2\, +\, 2ax\, +\, bx\, +\, a\, +\, b\, +\, c \right)\, -\, \left(ax^2\, +\, bx\, +\, c \right)\, =\,\)
. . . . .\(\displaystyle =\, ax^2\, -\, ax^2\, +\, bx\, -\, bx\, +\, c\, -\, c\, +\, 2ax\, +\, a\, +\, b\)
. . . . .\(\displaystyle =\, 2ax\, +\, a\, +\, b\)
Then show the subtraction of these results. You'll be left with a constant.
Have fun!
You could always show the algebra. For simplicity, use x - 1, x, and x + 1 for your three "typical" x-values. Then your three quadratic terms are:
. . . . .\(\displaystyle \mbox{first term: }\, a(x\, -\, 1)^2\, +\, b(x\, -\, 1)\, +\, c\, =\, ax^2\, -\, 2ax\, +\, bx\, +\, a\, -\, b\, +\, c\)
. . . . .\(\displaystyle \mbox{second term: }\, ax^2\, +\, bx\, +\, c\)
. . . . .\(\displaystyle \mbox{third term: }\, a(x\, +\, 1)^2\, +\, b(x\, +\, 1)\, +\, c\, =\, ax^2\, +\, 2ax\, +\, bx\, +\, a\, +\, b\, +\, c\)
Then subtract the first from the second:
. . . . .\(\displaystyle \left(ax^2\, +\, bx\, +\, c \right)\, -\, \left(ax^2\, -\, 2ax\, +\, bx\, +\, a\, -\, b\, +\, c \right)\)
. . . . .\(\displaystyle =\, ax^2\, -\, ax^2\, +\, bx\, -\, bx\, +\, c\, -\, c\, +\, 2ax\, -\, a\, +\, b\)
. . . . .\(\displaystyle =\, 2ax\, -\, a\, +\, b\)
...and the second from the third:
. . . . .\(\displaystyle \left(ax^2\, +\, 2ax\, +\, bx\, +\, a\, +\, b\, +\, c \right)\, -\, \left(ax^2\, +\, bx\, +\, c \right)\, =\,\)
. . . . .\(\displaystyle =\, ax^2\, -\, ax^2\, +\, bx\, -\, bx\, +\, c\, -\, c\, +\, 2ax\, +\, a\, +\, b\)
. . . . .\(\displaystyle =\, 2ax\, +\, a\, +\, b\)
Then show the subtraction of these results. You'll be left with a constant.
Have fun!
Okay! I will definitely show this. At least then I can show them algebraically why 2a = the 2nd difference.
Any other ideas out there? Is there a way to show the idea on a graph? If you graph the parabola, perhaps there is something about the width of the parabola? Does it have anything to do with the average slope?
Any other ideas out there? Is there a way to show the idea on a graph? If you graph the parabola, perhaps there is something about the width of the parabola? Does it have anything to do with the average slope?
mmm4444bot
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chouse said:
Try telling your students that this particular coefficient is one-half of the force due to gravitation because that's what experimental data show.
As soon as they object to this explanation, immediately invite them to gain deeper insight by becoming physicists.