How do you graph #y= 1/2x + 2#?

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Answer 1 · 2017-08-04T03:52:27

See a solution process below:

First, find two points on the line line:

For #x = 0#;

#y = (1/2 * 0) + 2#

#y = 0 + 2#

#y = 2# or #(0, 2)#

For #x = 4# (I picked #4# because it is easy to multiply it by #1/2#)

#y = (1/2 * 4) +2#

#y = 2 + 2#

#y = 4# or #(4, 4)#

Next plot these two points on the graph:

graph{(x^2 + (y-2)^2-0.125)((x-4)^2+(y-4)^2-0.125)=0 [-20, 20, -10, 10]}

Now, draw a line through the two points:

graph{(y-(x/2)-2)(x^2 + (y-2)^2-0.125)((x-4)^2+(y-4)^2-0.125)=0 [-20, 20, -10, 10]}