How do you graph #y<=3x+11# on the coordinate plane?
1 Answer
Sep 9, 2017
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:
For:
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]}