Question

How do you graph using slope and intercept of `6x - 12y = 24`?

Answer 1

Re-arrange the equation to get the base form of y=mx+b (slope-intercept form), build a table of points, then graph those points.

graph{0.5x-2 [-10, 10, -5, 5]}

Explanation:

The slope-intercept line equation is `y=mx+b`, where m is the slope and b is the point where the line intercepts the y-axis (a.k.a. the value of y when x=0)

To get there, we'll need to rearrange the starting equation some. First off is to move the 6x to the right-hand side of the equation. We'll do that by subtracting 6x from both sides:

`cancel(6x)-12y-cancel(6x)=24-6x rArr -12y=24-6x`

Next, we'll divide both sides by y's coefficient, -12:

`(cancel(-12)y)/cancel(-12)=24/(-12)-(6x)/(-12) rArr y=0.5x-2`

Now we have our slope intercept form of the equation, `y=0.5x-2`.

Next, let's build a table of points to plot. Since it's a straight line, we only need 2 points that we can line up with a ruler and draw a straight line through.

We already know one point, which is the y-intercept (0,-2). Let's pick another point, at `x=10`:

`y=0.5xx(10)-2`
`y=5-2 rArr y=3`

So our second point is (10,3). Now we can draw a straight line that passes through both of those points:

graph{0.5x-2 [-10, 10, -5, 5]}

Answer 2

`y=1/2x -2`

Explanation:

First you have to get the y by itself so you subtract 6x from both sides `-12y=24-6x`
Then, you want to get one y so you divide both sides by -12
`y=1/2x-2`
You then graph it so that the y-intercept is at -2, because at the y-intercept, x is always 0. And then you go up 1, over 2 every point after that.