Me out : ,( i’m just trying to graduate

(-6,-5)=> f(x)= x^2-2x-5

(8,11)=> f(x)=3x+5

(3,2)=> f(x)=4-x

(6,28)=> f(x)=(x+5) x^2

Step-by-step explanation:

For f(x)=3x+5

Putting the values of domain one by one:

Putting x=1:

f(x)=3(1)+5

=3+5

=8

Putting x = 2:

f(x)=3(2)+5

=6+5  

=11

For function f(x)= x^2-2x-5

Putting x=1:

f(x)=〖(1)〗^2-2(1)-5

=1-2-5

= -6

Putting x=2

f(x)=〖(2)〗^2-2(2)-5

=4-4-5

= -5

For f(x)=(x+5) x^2

Putting x=1  

f(x)=(1+5) (1)^2

=6*1

=6

Putting x=2  

f(x)=(2+5) (2)^2

=7*4

=28

For f(x)=4-x

Putting x = 1

f(x)=4-1

=3

Putting x=2

f(x)=4-2

=2

So the ranges belonging to the functions when domain is (1,2) are:

(-6,-5)=> f(x)= x^2-2x-5

(8,11)=> f(x)=3x+5

(3,2)=> f(x)=4-x

(6,28)=> f(x)=(x+5) x^2

the solution is 12x-6y