Lesson Objectives

- Demonstrate an understanding of how to graph a linear inequality in two variables
- Learn how to graph a second-degree inequality
- Learn how to graph a system of non-linear inequalities

## How to Graph a Second-Degree Inequality

A second-degree inequality is an inequality with at least one variable raised to the second power and no variable with an exponent larger than 2. When we graph a second-degree inequality, we use the following steps:

Example 1: Graph each inequality $$(x - 1)^2 + (y + 3)^2 < 25$$ We begin by graphing the boundary. We will replace the less than symbol "<" with an equality symbol "=": $$(x - 1)^2 + (y + 3)^2=25$$ This is the graph of a circle, with a center at (1,-3) and a radius of 5. Since we have a strict inequality, we will graph our boundary as a dashed circle. Now, we will choose a test point. If (0,0) is not on the boundary, it is usually the easiest to use. Let's plug a 0 in for x and a 0 in for y in the original inequality: $$(0 - 1)^2 + (0 + 3)^2 < 25$$ $$(-1)^2 + (3)^2 < 25$$ $$1 + 9 < 25$$ $$10 < 25 \require{color}\hspace{.25em}\color{green}{✔}$$ Since the test point works as a solution to the original inequality it lies in the solution region. Since (0,0) is inside of the boundary, we will shade everything inside of the boundary or circle.

- Graph the boundary, the boundary separates the solution region from the non-solution region
- To graph the boundary, replace the inequality symbol with an equality symbol. Graph the resulting equation.
- The boundary is solid for a non-strict inequality "≥" or "≤"
- The boundary is dashed or broken for a strict inequality ">" or "<"

- Use a test point to find and shade the solution region
- Recall that the test point can be any point that is not on the boundary
- If the test point works in the original inequality, the test point lies in the solution region
- If the test point fails in the original inequality, the test point lies in the non-solution region

Example 1: Graph each inequality $$(x - 1)^2 + (y + 3)^2 < 25$$ We begin by graphing the boundary. We will replace the less than symbol "<" with an equality symbol "=": $$(x - 1)^2 + (y + 3)^2=25$$ This is the graph of a circle, with a center at (1,-3) and a radius of 5. Since we have a strict inequality, we will graph our boundary as a dashed circle. Now, we will choose a test point. If (0,0) is not on the boundary, it is usually the easiest to use. Let's plug a 0 in for x and a 0 in for y in the original inequality: $$(0 - 1)^2 + (0 + 3)^2 < 25$$ $$(-1)^2 + (3)^2 < 25$$ $$1 + 9 < 25$$ $$10 < 25 \require{color}\hspace{.25em}\color{green}{✔}$$ Since the test point works as a solution to the original inequality it lies in the solution region. Since (0,0) is inside of the boundary, we will shade everything inside of the boundary or circle.

## How to Graph a System of Non-Linear Inequalities

When we encounter a system of non-linear inequalities, we will begin by graphing each inequality separately. Once this is done, we will shade the overlap between the graphs. This overlap is the section of the graph that satisfies all inequalities of the system. Let's look at an example.

Example 2: Graph each non-linear system of equations $$-2x + y ≤ -3$$ $$y ≥ x^2 - 3x - 4$$ To solve this system, we will graph each inequality separately. Let's begin by graphing our first inequality. $$-2x + y ≤ -3$$ Solve for y: $$y ≤ 2x - 3$$ We will replace the inequality symbol with an equality symbol. This will be our boundary line. $$y=2x - 3$$ Since we solved the inequality for y, we don't need to use a test point. Since y is less than or equal to 2x - 3, we know we need to shade below the line. Alternatively, if we used a test point of (0,0), we would see this point would lie in the non-solution region. $$-2(0) + (0) ≤ -3$$ $$0 ≤ -3 \hspace{.25em}\color{red}{✖}$$ Let's shade below the line: Now, let's think about our second inequality. $$y ≥ x^2 - 3x - 4$$ We will replace our inequality symbol with an equality symbol. This will give us our boundary. $$y=x^2 - 3x - 4$$ Vertex: (3/2, -25/4)

x-intercepts: (4,0),(-1,0)

y-intercept: (0,-4)

Additional Points: (3,-4), (5,6), (-2,6) Let's graph our boundary: If we use the test point of (0,0), we will see that it lies in the solution region. $$0 ≥ (0)^2 - 3(0) - 4$$ $$0 ≥ -4 \hspace{.25em}\color{green}{✔}$$ Now, let's shade the solution region. This will be inside of the parabola. Now, to find the solution for the system, let's overlay the two graphs on top of each other. The solution will be the overlap of the two solution regions. This area of the coordinate plane will satisfy both inequalities of the system. The area in orange is the overlap of the two graphs. We can complete our graph of the system by shading this area only:

Example 2: Graph each non-linear system of equations $$-2x + y ≤ -3$$ $$y ≥ x^2 - 3x - 4$$ To solve this system, we will graph each inequality separately. Let's begin by graphing our first inequality. $$-2x + y ≤ -3$$ Solve for y: $$y ≤ 2x - 3$$ We will replace the inequality symbol with an equality symbol. This will be our boundary line. $$y=2x - 3$$ Since we solved the inequality for y, we don't need to use a test point. Since y is less than or equal to 2x - 3, we know we need to shade below the line. Alternatively, if we used a test point of (0,0), we would see this point would lie in the non-solution region. $$-2(0) + (0) ≤ -3$$ $$0 ≤ -3 \hspace{.25em}\color{red}{✖}$$ Let's shade below the line: Now, let's think about our second inequality. $$y ≥ x^2 - 3x - 4$$ We will replace our inequality symbol with an equality symbol. This will give us our boundary. $$y=x^2 - 3x - 4$$ Vertex: (3/2, -25/4)

x-intercepts: (4,0),(-1,0)

y-intercept: (0,-4)

Additional Points: (3,-4), (5,6), (-2,6) Let's graph our boundary: If we use the test point of (0,0), we will see that it lies in the solution region. $$0 ≥ (0)^2 - 3(0) - 4$$ $$0 ≥ -4 \hspace{.25em}\color{green}{✔}$$ Now, let's shade the solution region. This will be inside of the parabola. Now, to find the solution for the system, let's overlay the two graphs on top of each other. The solution will be the overlap of the two solution regions. This area of the coordinate plane will satisfy both inequalities of the system. The area in orange is the overlap of the two graphs. We can complete our graph of the system by shading this area only:

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