How do you graph #y=3.00x+23#?

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Answer 1 · 2017-08-17T13:25:12

See a solution process below:

To graph a linear equation we need to plot two points on the line and then draw a line through them.

For #x = 0#

#y = (3.00 xx 0) + 23#

#y = 0 + 23#

#y = 23# or #(0, 23)#

For #x = 1#

#y = (3.00 xx 1) + 23#

#y = 3.00 + 23#

#y = 26# or #(1, 26)#

Next, we can plot the two points on the grid:

graph{(x^2+(y-23)^2-0.251)((x-1)^2+(y-26)^2-0.25)=0 [-35, 35, -5, 30]}

Now, we can draw the line through the two points:

graph{(x^2+(y-23)^2-0.251)((x-1)^2+(y-26)^2-0.25)(y-3x-23)=0 [-35, 35, -5, 30]}