Respuesta :
The function graph is a straight, so the function is the first-degree. The first-degree polynomial funciton of can be generalized as follows:
[tex]f(x)=ax+b[/tex]
To find the unkwnons, we need to extract two chart points. I chose these :(0;1), (14;8).
The slope (a) is give by:
[tex]a= \frac{\Delta y}{\Delta x} \\ a= \frac{(8-1)}{(14-0)} \\ a= \frac{1}{2} [/tex]
The linear correlation coefficient of the equation (b) is obtained when the variable is zero, therefore:
[tex]f(x)=ax+b \\ 1=0+b \\ b=1[/tex]
Throught it we have sought the equations is given by:
[tex]\boxed {f(x)= \frac{x}{2}+1} [/tex]
If you notice any mistake in my english, please let me know, because i am not native.
[tex]f(x)=ax+b[/tex]
To find the unkwnons, we need to extract two chart points. I chose these :(0;1), (14;8).
The slope (a) is give by:
[tex]a= \frac{\Delta y}{\Delta x} \\ a= \frac{(8-1)}{(14-0)} \\ a= \frac{1}{2} [/tex]
The linear correlation coefficient of the equation (b) is obtained when the variable is zero, therefore:
[tex]f(x)=ax+b \\ 1=0+b \\ b=1[/tex]
Throught it we have sought the equations is given by:
[tex]\boxed {f(x)= \frac{x}{2}+1} [/tex]
If you notice any mistake in my english, please let me know, because i am not native.