Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

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January 2019, Part II

All Questions in Part I are worth 2 credits. Partial credit can be earned.


25. Justify why (x²y⁵)^(1/3) / (x³y⁴)^(1/4) is equivalent to x^(-1/12)y^(2/3) using properties of rational exponents, where x =/= 0 and y =/= 0.

Answer:
As noted above in the way I had to type out the question, the cube root is the same as (1/3) power, and the fourth root is the same as (1/4) power.
Multiply the exponents:

(x^(2/3)y^(5/3)) / (x^(3/4)y^(4/4))
Next subtract the exponents of the two x terms and the two y terms
x^{(2/3) - (3/4)} y^{(5/3) - 1})
x^{(8/12) - (9/12)} y^{(5/3) - (3/3)}
x^(-1/12) y^(2/3)

26. The zeros of a quartic polynomial function are 2, -2, 4, and -4. Use the zeros to construct a possible sketch of the function, on the set of axes below.

Answer:
You needed to sketch something like what's below, with the end both pointing up or down. There should be three turning points (in this case 2 minimums and 1 maximum). Label the x-axis so that it's obvious that the zeroes are -4, -2, 2 and 4. The y-intercept isn't important but it should be a turning point.

Comments and questions welcome.

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