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

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Answer 1 · 2018-06-01T15:20:31

See a solution process below:

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

First Point: For #x = -1#

#y = (2 * -1) + 1#

#y = -2 + 1#

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

Second Point: For #x = 1#

#y = (2 * 1) + 1#

#y = 2 + 1#

#y = 3# or #(1, 3)#

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

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