Question

How do you graph the lines using slope-intercept form `y = -2/3x + 1`?

Answer 1

See a solution process below:

Explanation:

This equation is in slope intercept form. The slope-intercept form of a linear equation is: `y = color(red)(m)x + color(blue)(b)`

Where `color(red)(m)` is the slope and `color(blue)(b)` is the y-intercept value.

`y = color(red)(-2/3)x + color(blue)(1)`

Therefore, the y-intercept is: `color(blue)(b = 1)` or `(0, color(blue)(1))`

We can plot this point on the grid as:

graph{(x^2 + (y-1)^2 - 0.025) = 0 [-10, 10, -5, 5]}}

The slope is: `color(red)(m = -2/3)`

Slope is also: `"rise"/"run"`

In this case, the `"rise"` is `-2` so we need to go down 2 positions on the `y` value. And, the run is `3` so we need to go right 3 positions on the `x` value.

This second point is: `(0 + 3, 1 - 2) => (3, -1)`

We can now plot this point:

graph{(x^2 + (y - 1)^2 - 0.025)((x - 3)^2 + (y + 1)^2 - 0.025) = 0 [-10, 10, -5, 5]}}

Now, we can draw a line through the two points giving:

graph{(y + (2/3)x - 1)(x^2 + (y - 1)^2 - 0.025)((x - 3)^2 + (y + 1)^2 - 0.025) = 0 [-10, 10, -5, 5]}}