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

See a solution process below:

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-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]}

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To graph ( y \leq 3x + 11 ) on the coordinate plane, you first draw the line ( y = 3x + 11 ) as a dashed line (since it's not included in the inequality). Then, shade the region below the line to represent the inequality.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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