Algebra 2 Problems of the Day (open-ended)

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January 2018, Part II

Questions in Part II are worth 2 credits. All work must be shown or explained for full credit. A correct numerical answer without work is only worth 1 credit.


31.The zeros of a quartic polynomial function h are 1, +2, and 3.
Sketch a graph of y = h(x) on the grid below.

Answer:
They are asking for a "sketch", not an exact graph, because they did not give you the actual function. There are an infinite number of quartic (fourth power) graphs that have these zeroes.
You can enter the following into your graphing calculator to see an example:

y = (x + 2)(x + 1)(x - 2)(x - 3)
You can also multiply that by any coefficient, positive or negative. As always, a positive multiplier means that the graph opens upward, and a negative means that it will open downward. Either is acceptable.
Here is one possible answer:

Note that you needed to have the zeroes in the correct places for full credit. If you flipped the signs, for example, you would lose a credit.



32. 2 Explain why 81^3/4 equals 27.

Answer:

The fourth root of 81 is 3 and 3 to the third power is 27.
You can write it as 81 = (3 * 3 * 3 * 3)^3/4 = (3 * 3 * 3) = 27 and add an explanation.

Reminder "Explain" means that you have to write words to "explain" (sorry about the repetition, except that I'm not). "Justify" means that you can just write equations as proof, but "explain" means at least a sentence.



Comments and questions welcome.

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