How do you graph #y=0.25x-2# using a table of values?

1 Answer
Jan 27, 2016

Generate a table of #(x,y)# coordinates using several (arbitrary, but convenient) values for #x#;
plot these coordinates;
connect the plotted points with a line.

Explanation:

To simplify the coordinates, I would suggest using using values of #x# which are multiples of #4# (so #0.25x# will an integer).

Here are some possibilities:

#{: (color(red)(x)," | ",color(blue)(y=0.25x-2)), (bar(color(white)("XXXX")),,bar(color(white)("XXXXXXXXXX"))), (-4," | ",-3), (color(white)("X")0," | ",-2), (color(white)("X")4," | ",-1), (color(white)("X")8," | ",color(white)("X")0), (12," | ",color(white)("X")1) :}#

Plotting the points:
graph{((x+4)^2+(y+3)^2-0.02)(x^2+(y+2)^2-0.02)((x-4)^2+(y+1)^2-0.02)((x-8)^2+y^2-0.02)((x-12)^2+(y-1)^2-0.02)=0 [-9.35, 15.96, -6.85, 5.81]}

Connected with a line:
graph{((x+4)^2+(y+3)^2-0.02)(x^2+(y+2)^2-0.02)((x-4)^2+(y+1)^2-0.02)((x-8)^2+y^2-0.02)((x-12)^2+(y-1)^2-0.02)(0.25*x-2-y)=0 [-9.35, 15.96, -6.85, 5.81]}