How do you graph #y= 2x+2# by plotting points?

1 Answer
MathFact-orials.blogspot.com
Apr 14, 2018

See below:

Explanation:

My favourite way to plot graphs is to put the equation into slope-intercept form (which our question is) and then plot from there.

The slope-intercept form, in general, is:

#y=mx+b#, where #m=" slope" and b= y-"interecept"#

The first point we're given is the #y# intercept, which is #(0,2)#.

We're also told the slope, which is in the form of #"rise"/"run"#, or in other words, for every number of steps we take up (the numerator) we take some number of steps to the right. Our slope is 2, which means we move up 2 for every 1 we move right. We can then find our next point by saying:

#(0+1, 2+2)=>(1,4)#

Let's graph those 2 points and connect them with a line:

graph{(y-2x-2)((x-0)^2+(y-2)^2-.1)((x-1)^2+(y-4)^2-.1)=0}

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