Graphing Absolute Value Inequalities: Find The Non-Solution

Hey guys! Today, we're diving into the awesome world of graphing inequalities, specifically those involving absolute value. It's a super useful skill in math, and honestly, it's not as scary as it might sound at first. We're going to tackle a specific problem: graphing the inequality $y \leq|x+2|$ and then figuring out which point isn't part of the solution. Grab your notebooks, maybe a snack, and let's get this done!

Understanding Absolute Value Inequalities

First things first, let's break down what an absolute value inequality actually is. When you see that vertical bar notation, like $|x+2|$, it means the distance of whatever's inside from zero. For example, $|5| = 5$ and $|-5| = 5$. The absolute value is always non-negative. Now, when we combine this with an inequality sign (like $\leq$, $\geq$, $<$, or $>$) and a variable like $y$, we're looking at a whole region on a graph, not just a single line. Our specific inequality, $y \leq|x+2|$, means we're looking for all the points $(x, y)$ where the $y$-coordinate is less than or equal to the absolute value of $(x+2)$.

To graph $y = |x+2|$, we first think about the parent function $y = |x|$. This graph looks like a 'V' shape with its vertex at the origin $(0,0)$. The absolute value bars basically mean that whatever value $x$ is, the output $y$ will always be positive or zero. Now, our inequality has $(x+2)$ inside the absolute value. This is a horizontal shift. Just like with regular functions, adding a number inside the parentheses shifts the graph to the left. So, $y = |x+2|$ will have its vertex shifted 2 units to the left, meaning its vertex is at $(-2, 0)$. From this vertex, the graph will go up and to the right with a slope of 1, and up and to the left with a slope of -1. It forms that classic 'V' shape, but centered at $(-2, 0)$.

Now, for the inequality part: $y \leq|x+2|$. The '$\|y aleq$' sign tells us we need to shade the region below the line (or in this case, below the 'V' shape graph of $y = |x+2|$). The 'equal to' part ($\leq$) means the boundary line itself is included in the solution. So, we draw a solid line for $y = |x+2|$. If it were $y < |x+2|$, we'd draw a dashed line to show the boundary isn't included. So, the solution to $y \leq|x+2|$ is all the points on the 'V' shape graph and everything below it.

Graphing $y

leq|x+2|$

Let's get this graphed, guys. First, we'll plot the boundary line $y = |x+2|$. We know the vertex is at $(-2, 0)$. Let's find a couple more points to get the 'V' shape.

• If $x = 0$, then $y = |0+2| = |2| = 2$. So, $(0, 2)$ is on the graph.
• If $x = -1$, then $y = |-1+2| = |1| = 1$. So, $(-1, 1)$ is on the graph.
• If $x = -3$, then $y = |-3+2| = |-1| = 1$. So, $(-3, 1)$ is on the graph.
• If $x = -4$, then $y = |-4+2| = |-2| = 2$. So, $(-4, 2)$ is on the graph.

We connect these points to form a solid 'V' shape with the vertex at $(-2, 0)$. The sides of the 'V' extend upwards indefinitely.

Now, for the shading part. Since our inequality is $y \leq|x+2|$, we need to shade the region below this 'V' shape. To test a region, we can pick a point that is not on the boundary line. A super easy point to test is usually the origin $(0, 0)$, if it's not on the boundary. Let's plug $(0,0)$ into our inequality: Is $0 leq |0+2|$? Is $0 leq |2|$? Is $0 leq 2$? Yes, it is! Since $(0, 0)$ makes the inequality true, we shade the region that contains $(0, 0)$. In this case, $(0, 0)$ is below the right arm of the 'V', so we shade all the area below the 'V' shape.

The solution set includes all points on the solid 'V' boundary and all points in the shaded region below it. Any point $(x, y)$ that falls on this solid boundary or within the shaded area satisfies the inequality $y leq|x+2|$.

Determining Which Point Is NOT Part of the Solution

Okay, so we've got our graph. The question asks us to identify which of the given points is NOT part of the solution. This means we're looking for a point that does not satisfy the inequality $y leq|x+2|$. To do this, we can either visually inspect our graph or, more reliably, plug each point's coordinates into the inequality and see if it holds true. If it doesn't hold true, that's our answer!

Let's test each point:

A. $(1, 2)$ Plug $x=1$ and $y=2$ into $y leq|x+2|$: $2 leq |1+2|$ $2 leq |3|$ $2 leq 3$ This statement is TRUE. So, $(1, 2)$ is part of the solution.

B. $(-1, 2)$ Plug $x=-1$ and $y=2$ into $y leq|x+2|$: $2 leq |-1+2|$ $2 leq |1|$ $2 leq 1$ This statement is FALSE. Since the inequality is false for this point, $(-1, 2)$ is NOT part of the solution. Bingo!

C. $(-1, -2)$ Plug $x=-1$ and $y=-2$ into $y leq|x+2|$: $-2 leq |-1+2|$ $-2 leq |1|$ $-2 leq 1$ This statement is TRUE. So, $(-1, -2)$ is part of the solution. (Notice this point is in the shaded region below the vertex).

D. $(0, 0)$ Plug $x=0$ and $y=0$ into $y leq|x+2|$: $0 leq |0+2|$ $0 leq |2|$ $0 leq 2$ This statement is TRUE. So, $(0, 0)$ is part of the solution. We actually used this point to help us determine the shading earlier!

Conclusion

So, after plugging in all the coordinates, we found that the point $(-1, 2)$ does not satisfy the inequality $y leq|x+2|$. This means it lies outside the shaded region and above the boundary line $y = |x+2|$. It's super important to double-check your calculations when testing points, as a small arithmetic error can lead you to the wrong conclusion. Remember, graphing inequalities gives you a visual representation of all possible solutions, and testing points is the definitive way to confirm whether a specific point is included or excluded from that solution set. Keep practicing these, guys, and you'll be graphing absolute value inequalities like a pro in no time! What other math mysteries can we solve together?