Question

How do you graph `y<=3x+11` on the coordinate plane?

Answer 1

See a solution process below:

Explanation:

First, solve for two points as an equation instead of an inequality to find the boundary line for the inequality.

For: `x = 0`

`y = (3 * 0) + 11`
`y = 0 + 11`
`y = 11` or `(0, 11)`

For: `x = 1`

`y = (3 * 1) + 11`
`y = 3 + 11`
`y = 14` or `(1, 14)`

We can now graph the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.

The boundary line will be solid because the inequality operator contains an "or equal to" clause.

graph{(x^2+(y-11)^2-0.125)((x-1)^2+(y-14)^2-0.125)(y-3x-11)=0 [-30, 30, -10, 20]}

Now, we can shade the right side of the line.

graph{(y-3x-11) <= 0 [-30, 30, -10, 20]}