In Exercises 27–62, graph the solution set of each system of ineq...

Question

In Exercises 27–62, graph the solution set of each system of inequalities or indicate that the system has no solution. x^2+y^2≤1, y−x^2>0

Graph showing the solution set for the inequalities x^2 + y^2 ≤ 16 and y - x^2 > -4.

Accepted Answer

Hello, today we're gonna be solving the system of inequalities by using the method of graphing. So the system of inequalities given to us is a system of two inequalities. We are given X squared plus Y squared is less than or equal to 16. And we are given Y minus X squared is greater than negative four. So the best way to solve this system of inequalities is to first graph the or the individual inequalities. So let's take a look at the first inequality. We are given X squared plus Y squared is less than or equal to 16. Now, if we replace the inequality sign with an equal sign, we get X squared plus Y squared is equal to 16. This matches the equation of a circle. And this is the equation of a circle where R squared is equal to 16. If we solve for R squared by taking the square root, we get R is equal to four. So that means that our inequality represents the equation of a circle with a radius of four. So if we were to graph that we have the equation of a circle where the end points are located at four, comma 00, comma four negative four, comma zero and zero, comma negative four. Now the question is do we want to include any values that lie on the actual graph of the circle while the inequality is using less than or equal to symbol? So that means that since we have the equal part, we can include any values that lie on the graph of the circle. So we're gonna graph this using a solid line. Next, we want to figure out what regional solutions we want to include in our graph. Do we want to include the region that lies outside of the circle or do we want to include the region that lies inside of the circle? While the inequality is saying that any X and Y values that we square and add together must be less than the radius and those X and Y values less than the radius are gonna be contained inside of the circle. So that means we want to include the regional solutions that lie inside of the circle. And this is going to be the graph for the first inequality. Now let's take a look at the second inequality. The second inequality is given to us as Y minus X squared is greater than negative four. If we were to replace the inequality sign with an equal sign, we get Y minus X squared is equal to negative four. If we solve for the Y variable by adding X squared to both sides, we get Y is equal to X squared minus four. This matches the equation of a Parabola. So if we rewrite our inequality by solving for the Y variable, we get Y is greater than X squared minus four. This is going to be the equation. This is going to match the equation of a Parabola where the center is located at zero comma negative four. But to get an accurate Parabola, let's go ahead and solve for the X intercepts, the X intercepts are going to exist where Y is equal to zero. So we have zero is equal to X squared minus four. If we add four to both sides of the equation, we get X squared as equal to positive four and taking the square root of both sides gives us X is equal to plus or minus two. So we're going to have two X intercepts one at two comma zero and another one at negative two comma zero. So since we have the two X intercepts and the center or the, at least the vertex of the Parabola, we can go ahead and graph for inequality. So again, the inequality is gonna be represented by a Parabola whose vertex is at zero, comma negative four and who has X intercepts at two comma zero and negative two comma zero. Now again, do we want to graph this Parabola using a solid or dotted line while the inequality is using the greater than symbol. And since we don't have the equal two part, that means we don't want to include any values that lie on the graph of the Parabola. So we're going to graph the Parabola using a dotted line. Now again, do we wanna to graph the region that is contained inside of the Parabola or do we want to include the region that is outside the Parabola? While the easiest way to determine this is to take a testing point, if we take the test, testing 0.0 comma zero and plug it into our inequality and the inequality becomes a true inequality, then we want to include the solution set that is contained inside the Parabola. So let's go ahead and plug in 00 into our inequality that is going to give us zero is greater than zero squared, which is zero minus four, zero minus four is going to give us negative four and we have zero is greater than negative four, which is a true statement. So since the testing point made this a true statement, we're going to want to include the solutions that are contained inside of the Parabola. Now that we know what the shape of each of the graphs look like. Let's go ahead and combine them into the same axis. So again, the first inequality was the graph of a circle with radius four and this is a graph of a circle with a solid line. Next we know that the equation or the inequality. The second inequality is represents a parabola whose vertex is at zero, comma negative four and who has X intercepts at two comma zero and negative two comma zero. And we graph this parabola using a dotted line. Finally, the solution set to the equate to the system of inequalities is going to be the overlapping section of the shaded region. So for the circle, we know that the entirety of the circle is shaded. But the inside portion of the Parabola is also shaded, which means that this region between the circle and the Parabola is going to be the solution set to the system of inequalities. So this is going to be the solution to the system of inequalities. Now we just need to select the graph that accurately represents this information. And this graph is going to be option C see option C gives us the circle with a radius of four and it gives us the Parabola with the dotted line and X intercepts at two comma zero and negative two comma zero. And the intersecting region of the two graphs is going to be this cone shape. So with that being said, the answer to this problem is going to be C so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 27–62, graph the solution set of each system of inequalities or indicate that the system has no solution. x^2+y^2≤1, y−x^2>0

Key Concepts

Inequalities

Graphing Systems of Inequalities

Solution Set

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