PQR with P(3, 6), 2-6, 6), R(0, 9); (x, y) = (x-3, y + 7) оа
Ob
P'(0, 13), O'(-9, 13), R'(-3, 16)
P'(10,3), Q'(1,3), R' (7, 6)
P'(3, 7), e'(-6, 7), R' (0,7)
P'(-3, 13), e' (3, 13), R'(-3, 16)

We have the answer as:

P'(0, 13), O'(-9, 13), R'(-3, 16)

Step-by-step explanation:

Here, we want to find the coordinates of the image after applying the transformation rule to the pre-image

what the transformation rule is saying is that we subtract 3 from the x-coordinate and add 7 to the y-coordinate

mathematically, we have these as follows;

P’ = (3-3, 6 + 7) = (0,13)

Q’ = (-6-3, 6 + 7) = (-9,13)

R’ ( 0-3, 9 + 7) = (-3, 16)

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.

this is a good question! i'll do some research and come back when i have the answer!