Aline is drawn through (–4, 3) and (4, 3). which describes whether or not the line represents a direct variation?

the line represents a direct variation because = negative startfraction 4 over 3 endfraction equals startfraction 4 over 3 endfraction.
the line represents a direct variation because it is horizontal.
the line does not represent a direct variation because it does not go through the origin.
the line does not represent a direct variation because –4(3) ≠ 4(3).

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.

the first domain and range set to the left is matched with the top relation, and the second set the set in the middle goes with the bottom relation.