Which is true about the solution to the system of inequalities shown? y > 3x + 1 y < 3x – 3 On a coordinate plane, 2 solid straight lines are shown. The first line has a positive slope and goes through (negative 2, negative 5) and (0, 1). Everything to the left of the line is shaded. The second line has a positive slope and goes through (0, negative 3) and (1, 0). Everything to the right of the line is shaded. Options:

Only values that satisfy y > 3x + 1 are solutions.

Only values that satisfy y < 3x – 3 are solutions.

Values that satisfy either y > 3x + 1 or y < 3x – 3 are solutions.

There are no solutions.

please help :(

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.

tried on you are knowing that we can't leave the picture in the graph ok no problem now you need to send it in the question group