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

##### 1 Answer

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

We're now in slope-intercept form, where

And so in our question,

Let's first graph the

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

Now let's plot a second point.

Our

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}