The graphs of the functions f(x) = va and g(x) = x - 2 are shown in the xy-coordinate plane. When the equation v = x - 2 is solved by squaring both sides of the equation, the algebraic solutions to the squared
equation are x = 1 and x = 4. What do the graph of f and g reveal about the solutions?
8
y = f(x)
2),
y = g(x)
A) Both 1 and 4 are solutions
B) 4 is a solution but l is not
C) geither 1 nor 4 is a solution
D) there are two extraneous solutions

nothjing

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.

4+3x hope you like mark as brainliest✔

this "question" looks more like a project is there any other info you can give me about this. it could be too complex for someone else to do on this website