Question

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

Answer 1

graph{y=-x+3/2 [-10, 10, -5.21, 5.21]}
This is the line that you should get from graphing `y=-x+3/2`

Explanation:

The format of this equation is `y=mx+b`, where `m` is the slope and `b` is the `y`-intercept. Given this, since we know that `b=3/2`, the graph begins at point `(0, 3/2)`, `3/2` above the origin `(0,0)`. From here, since the slope is `-1`, then you can just
begin counting down one, and right one point. From there, you should create a graph similar to the one shown in the answer.

Helpful Sources:

https://weteachscience.org/mentoring/resources/lesson-plans/algebra-1-%E2%80%93-how-to-graph-a-linear-equation-using-slope-and-y

https://www.khanacademy.org/math/algebra/two-var-linear-equations/graphing-slope-intercept-equations/v/graphing-a-line-in-slope-intercept-form

Answer 2

See Explanation

Explanation:

The function `y=mx+b` is defined as the equation of a line where `m` is the slope and `b` is the `y`-intercept.

In the function you gave

`y=-x+3/2`

The slope is `-1` and the `y`-intercept is `3/2`

To graph this, you begin at the point `(0,3/2)` because this is your `y`-intercept. From there, your slope which is `"rise"/"run"` which is `-1` means you go down `1` unit and then right `1` unit.

Essentially, the graph looks like this: (You can interact with the graph to get the exact points to plot)

graph{-x+3/2 [-10, 10, -5, 5]}