Ineed
a triangle has vertices at b(−3, 0), c(2, −1), d(−1, 2). which transformation would produce an image with vertices b″(1, −2), c″(0, 3), d″(3, 0)?

(x, y) → (x + 1, y + 1), (x, y) → (y, x)
(x, y) → (x + 1, y + 1), (x, y) → (−x, y),
(x, y) → (x, −y), (x, y) → (x + 2, y + 2)
(x, y) → (−x, y), (x, y) → (x + 2, y + 2)

answer: both chords and diameters have two endpoints on a circle. diameters must intersect the center of a circle.

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.

i think its doubled. let me know

step-by-step explanation: