What set of transformations are applied to parallelogram ABCD to create A'B'C'D'? Parallelogram formed by ordered pairs A at negative 4, 1, B at negative 3, 2, C at negative 1, 2, D at negative 2, 1. Second pa

(4 points)

Reflected over the x-axis and rotated 180°
Reflected over the y-axis and rotated 180°
Reflected over the x-axis and rotated 90° counterclockwise
Reflected over the y-axis and rotated 90° counterclockwise

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.

1332.

this is because to get from 333 to 999 you multiply by 3, and you should do the same to the number of hours.

therefore you would do 444*3=1332