How do you graph #y=1/3x+2#?

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Answer 1 · 2018-03-28T11:47:56

See a solution process below:

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

First Point: For #x = 0#

#y = (1/3 * 0) + 2#

#y = 0 + 2#

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

Second Point: For #x = 3#

#y = (1/3 * 3) + 2#

#y = 1 + 2#

#y = 3# or #(3, 3)#

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

graph{(x^2+(y-2)^2-0.035)((x-3)^2+(y-3)^2-0.035)=0 [-10, 10, -5, 5]}

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

graph{(y - (1/3)x - 2)(x^2+(y-2)^2-0.035)((x-3)^2+(y-3)^2-0.035)=0 [-10, 10, -5, 5]}