Help please and thanks! I'll give brainliest as well, but show your work please! (Or step by step :)

Step-by-step explanation:

<QPR = <RST(given)

<PRQ = <TRS(vertically opposite angles)

<PQR = < STR (remaining angles)

triangles PQR~TRS (by A.A.A. axiom)

now

PR/RS = PQ/ST=QR/RT (corresponding sides of similar triangles are proportional)

or, 15/6=PQ/(2x-4)=x/RT

or, 15/6 = = PQ.RT=x.(2x-4)

or,, 15/6 = 2x²-4x

or, 15= 12x²-24x

or, 12x²-24x-15 = 0

or, 12x²-(30-6)x-15=0

or, 12x²-30x+6x-15=0

or, 2x(6x-15)+1(6x-15)=0

or, (6x-15)(2x+1)=0

Either,

6x-15 = 0

or, 6x = 15

so, x = 15/6

so, x = 2.5

or,

2x+1=0

2x=-1

or, x = -1/2

Value of x is either 2.5 or -1/2

the answer is y = 11.54

hope this : )

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.