How do you graph #y=4+2/3x# on a coordinate grid?

Question

1997 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-16T11:53:57

See a solution process below:

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

First Point: For #x = 0#

#y = 4 + (2/3 * 0)#

#y = 4 + 0#

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

Second Point: For #x = 3#

#y = 4 + (2/3 * 3)#

#y = 4 + 6/3#

#y = 4 + 2#

#y = 6# or #(3, 6)#

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

graph{(x^2+(y-4)^2-0.1)((x-3)^2+(y-6)^2-0.1)=0 [-20, 20, -10, 10]}

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

graph{(y - 4 - (2/3)x)(x^2+(y-4)^2-0.1)((x-3)^2+(y-6)^2-0.1)=0 [-20, 20, -10, 10]}