The coordinates of the vertices of △jkl are j(1, 4) , k(6, 4) , and l(1, 1) .

the coordinates of the vertices of △j′k′l′ are j′(0, −4) , k′(−5, −4) , and l′(0, −1) .

what is the sequence of transformations that maps △jkl to △j′k′l′ ?

drag and drop the answers into the boxes to correctly complete the statement.

a sequence of transformations that maps △jkl to △j′k′l′ is a followed by a .

answer: my brother can figure it out, imma go get him so i can answer : ) edit: my brother is sleeping, sorry man

step-by-step explanation:

idk wbu

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.