How do you graph the inequality #y<=-4x+12#?

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

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-12)^2-0.25)((x-3)^2+y^2-0.25)(y+4x-12)=0 [-30, 30, -15, 15]}

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

graph{(y+4x-12) <= 0 [-30, 30, -15, 15]}

To graph the inequality ( y \leq -4x + 12 ), follow these steps:

1. Begin by graphing the line ( y = -4x + 12 ) as if it were an equation.
2. Since the inequality includes the "less than or equal to" symbol ( (\leq) ), the solution also includes the points on the line itself.
3. To determine which side of the line to shade, choose a test point not on the line. A common choice is the origin (0,0).
4. Substitute the coordinates of the test point into the original inequality. If it satisfies the inequality, shade the region containing the test point; if not, shade the opposite side.
5. In this case, if we substitute ( x = 0 ) and ( y = 0 ) into ( y \leq -4x + 12 ), we get ( 0 \leq -4(0) + 12 ), which simplifies to ( 0 \leq 12 ), indicating that the origin satisfies the inequality.
6. Therefore, shade the region below the line ( y = -4x + 12 ) to represent the solution set of the inequality.