Question

How do you graph using slope and intercept of `2x+3y= -1`?

Answer 1

See below

Explanation:

The first thing I'd do is to change the equation from the current standard form and put it into slope-intercept form. We do that by solving for `y`:

`2x+3y=-1`

`3y=-2x-1`

`y=-2/3x-1/3`

We're now in slope-intercept form, where

`y=mx+b, m="slope" and b = y-"intercept"`

And so in our question, `m=-2/3` and `b=-1/3`

Let's first graph the `y`-intercept. That's at `(0,-1/3)`:

graph{((x-0)^2+(y+1/3)^2-.1)=0}

Now let's plot a second point.

`m=(Delta y)/(Delta x)="rise"/"run"`

Our `m=-2/3`. For every 2 that we move up, we move 3 to the left (I'm dealing with the negative sign by having us move left - with a positive slope we'd move to the right). We can start from our first point and move in that way, and so our second point can be found by writing:

`(0-3, -1/3+2)=(-3,5/3)`

Let's plot that:

graph{((x-0)^2+(y+1/3)^2-.1)((x+3)^2+(y-5/3)^2-.1)=0}

And now connect the two points with a line:

graph{((x-0)^2+(y+1/3)^2-.1)((x+3)^2+(y-5/3)^2-.1)(2x+3y+1)=0}