How do you graph #y=1/2x+1/4#?

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Answer 1 · 2017-08-26T14:47:28

See a solution process below:

First, solve for two points which solve the equation and plot these points:

First Point:

For #x = 0#

#y = (1/2 * 0) + 1/4#

#y = 0 + 1/4#

#y = 1/4# or #(0, 1/4)#

Second Point:

For #x = 1/2#

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

#y = 1/4 + 1/4#

#y = 2/4#

#y = 1/2# or#(1/2, 1/2)

We can next graph the two points on the coordinate plane:

graph{(x^2+(y-0.25)^2-0.00075)((x-0.5)^2+(y-0.5)^2-0.00075)=0 [-2, 2, -1, 1]}

Now, we can draw a straight line through the two points to graph the line:

graph{(y-0.5x-0.25)(x^2+(y-0.25)^2-0.00075)((x-0.5)^2+(y-0.5)^2-0.00075)=0 [-2, 2, -1, 1]}