We are asked to graph the functions C(x)=12x+17000 and C(x)=20x+16000.
See the attachment for the graphs. C(x)=12x+17000 is in green while C(x)=20x+16000 is in blue.
C(x)=12x+17000 is a linear function in x. The independent variable is x. We can rewrite the equation as a linear function of two variables, x and y, where y=C(x) is the dependent variable. Thus, y=12x+17000.
As x is the independent variable, you can choose any real number for x. However, you do have to take into consideration the domain of the function. For example if C(x) is the cost related to producing x units, then C(x) is greater than or equal to zero. And there is probably some upper limit on the number of units you can produce for a set cost—if nothing else, the cost of raw materials would increase as you use up most of the readily available material.
For this function, we see that the y-intercept is 17000 and the slope (rate of change) is 12. If C(x) is the cost of producing x units, then the initial cost (raw materials, equipment, etc.) is 17000 while the unit cost is 12.
To graph, we can choose any real number for x in the domain. It is reasonable to choose convenient values for x. Here, as x increases by 50 units, the function increases by 600; or change x by 100 units, and the y increases by 1200. For the graph in the attachment, I chose x=0,100,200,300,400,500, etc. You then substitute the chosen value of x into the function. If x=100, then y=C(x)=12(100)+17000=18200, etc.
This produces the points (0,17000), (100,18200), (200,19400), (300,20600), etc. You only need two points to draw a line. You can rescale the dependent variable axis as needed—make sure to label the units.
** From the graph, and algebraically confirmed, the two graphs intersect at (125,18500). Under the assumption we are looking at cost functions after producing 125 units it would be cheaper to choose the green option (12x+17000) despite the higher initial outlay. If you plan to make 125 or less units, the better (cheaper) option would be the blue (20x+16000.)