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

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Answer 1 · 2017-10-17T12:31:04

See a solution process below:

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

First Point: For #x = 0#

#y = 0 xx 1/4#

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

Second Point: For #x = 4#

#y = 4 xx 1/4#

#y = 4/4#

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

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

graph{(x^2+y^2-0.025)((x-4)^2+(y-1)^2-0.025)=0 [-10, 10, -5, 5]}

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

graph{(y - (1/4)x)(x^2+y^2-0.025)((x-4)^2+(y-1)^2-0.025)=0 [-10, 10, -5, 5]}