Discrete math

which of the following functions f : {0, 1, 2, 3} ! {0, 1, 2, 3} are onto?
1) f(x) = x
2) f(x) = x^2 mod 4
3) f(x) = x^2 - x mod 4

4) f(0) = 3, f(1) = 2, f(2) = 1, f(3) = 0
5) f(0) = 1, f(1) = 2, f(2) = 1, f(3) = 2

1) yes

2) No

3) No

4) yes

5) No

Step-by-step explanation:

1) f(0)=0, f(1)=1, f(2)=2, f(3)=3, then

2) f(0)=0, , then f(1)=1; , then f(2)=0; , then f(3)=1

Then , this means that f isn't onto.

3.

Then  , this means that f isn't onto.

4.  , then f is onto.

5. , this means that f isn't onto.

the final function for the graph becomes x-2=-1. before graphing you solve the equation, where x=1. then graph accordingly.   ( on the x-axis, 1 on the line will possess a vertical line.)

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option