Question 7 of 15 Miguel Is reflecting a point (-5, 7) across the lines y = x, y = -x. y=0, and I = 0. let Q' be the image of
Qarze
after each
reflection
Which statements are true? Select all that apply.
For y=x, the image of point Q will be Q'[(7,5))
For x=0, the image of point Q will be Q1/5, 7)).
Asses
For y=x, the image of point Q will be Q'UZ. -5)).
For y=-x, the image of point Q will be Q'((5.7)).
Stand
For y=0, the image of point Q will be Q'l(-5, -7)).
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i do not know this sorry : (

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i cant fully remember all of the ranks, here are the ones i remember.

hand, ambitous, ace, genius.

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.