Which table shows a function that is decreasing only over the interval (–1, 1)? A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 0, 3, 0, negative 3, 0. A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 10, 8, 0, negative 8, negative 10. A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 0, negative 3, 0, 3, 0. A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 10, negative 8, 0, 8, 10.

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step-by-step explanation:

i need more details

step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.