How do you graph #y=3x+4#?

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Answer 1 · 2017-10-24T19:38:22

See a solution process below:

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

First Point: For #x = 0#

#y = (3 * 0) + 4#

#y = 0 + 4#

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

First Point: For #x = 1#

#y = (3 * 1) + 4#

#y = 3 + 4#

#y = 7# or #(1, 7)#

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

graph{(x^2+(y-4)^2-0.1)((x-1)^2+(y-7)^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-3x-4)(x^2+(y-4)^2-0.1)((x-1)^2+(y-7)^2-0.1)=0 [-20, 20, -10, 10]}