Question

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

Answer 1

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}