Graph the solution set of each system of inequalities.
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Question

Graph the solution set of each system of inequalities.

y ≥ x² + 4x + 4

y < -x²

Accepted Answer

Hello everybody. I have everything right today, today we're going to begin this question that states provide the graph of the solution set of the following system where our system includes two equations. And we'll have equation number one being Y is greater than, or equal to X squared plus eight, X plus 19. And equation number two consists of Y is less than negative open parentheses, X minus one closed parentheses squared. Now, the first thing I'll note in these two inequalities is I'll note that inequality number one is a non strict inequality and this is because it is greater than or equal to. So because it's or equal to, it is a non strict inequality and we will therefore have a solid graph. So a solid line graph and if I look at in quantity, number two, I will note that this is a strict inequality. So this is less than there is no or equal to. So it is a strict inequality and we will therefore have a dashed graph. And this will be more clear once we actually get to the graphing part. So the first thing I'll do here now is we have two inequalities. So let's tackle each 11 by one. So I'll focus on inequality number one first. And the first thing I'll do is change the greater than, or equal to sine to an equal sign. So have Y equals X squared plus eight X plus 19. And now another thing that I'm going to do is I have this in a form kind of in that we could factor, we have X squared plus eight X plus 19. But I, what I want to do is complete the square. So what that means is basically, I'm going to find a way to turn this into a perfect square. Now, to do that, I know I know a version of this equation that's not far off. We have Y equals open parentheses, X squared plus eight X. Now, if it was plus 16, I know that that could be a perfect square. So I'll have plus 16 and close that parentheses and then I'll have plus open parentheses, 19 minus 16, closed parentheses. And this would be 19 minus 16 is three. Now, I imagine you had the 16 now that we have X squared plus eight X plus 16, if we were to carry out all this, these equations will have the 16 at the end plus 19 minus 16, which is three. So we, and then we would be 16 plus two, which is 19 and we would be back to our X squared plus eight X plus 19. So all we did was divide the 19 as a 19 minus 16 and a 16 to have a perfect square. So if we have a perfect square, we're gonna have now, Y equals open parentheses, X plus four, closed parentheses, squared plus three. And now we can turn this into Y equals open bracket, X minus open parentheses, negative four, closed parentheses, close bracket squared plus three. So this is just a different way of writing the X plus four. So if we have X minus negative four, we have X plus four. So it's just a different way of writing that. And we have that so that now we can look at the vertex form of quadratic function. So the reason why I changed the plus four to a minus negative four, it is because now if we look at the vertex form of a quadratic function, we have that Y equals a multiplying X minus H closed parentheses squared plus K where open parentheses, H comma K closed parentheses is the vertex. So by changing our formula to have X minus negative four, I can have it in the form in the vertex form. So I can see that therefore we will have I were vertex is negative four comma three because the plus three is the K and the negative four is the H. So now we have the vertex of inequality number one and now we can plot some points. So basically this is just to test different points, see where they would lie on our graph. In this case, I'll test X equals negative five. So what you do here is basically just plug in negative five for the equation that we had for the X and the equation that we had. So we have Y equals open bracket, open parentheses, negative five, closed parentheses minus open parentheses, negative four closed parentheses, closed bracket squared plus three. And this means that Y equals negative one squared plus three, which is the same as saying one plus three because negative one squared is one. So we have that Y equals four or the point negative five comma four. So now we have a vertex and we have one point and now we'll test a different point, let's say X equals negative three and you can test any point you want, these are just specific points that I'll be testing. So if we have X equal to negative three, we'll have that Y equals open bracket, open parentheses, negative three, closed parentheses minus open parentheses, negative four, closed parentheses, closed bracket squared plus three, which means that Y equals positive one squared plus three, which means that Y equals four or the point negative three comma four. So now we have a second point for our inequality. And now, because this is an in inequality, we know we have to test different points in our graph. So that means because we have an inequality, we know there will be multiple answers. So in the case of inequality, number one, we have that Y E Y is greater than, or equal to. So the answer to that would be anything that is greater than, or equal to that Y so they will meet multiple answers. So let's test, I'm going to test the origin. You can test any point that you want but I will be testing zero, comma zero and you do this because since you have multiple answers, you have to shade a region of your graph representing those multiple answers. So we'll, we'll see if the origin falls within that region. So if we test zero comma zero, we have that zero is greater than or equal to zero squared plus eight multiplied by zero plus 19, which means that zero is greater than or equal to zero. Squared is zero plus eight times zero is zero. So zero plus zero is zero plus 19 is 19. So we have that zero is greater than or equal to 19, which is false, which means that the origin does not fall within the region that we will be shading. And that will be it for equation number one and we can move on to equation number two. So we have the same thing we'll do first is substitute the last 10 sign with an equal sign. So we'll have Y is equal to negative open parentheses, X minus one, closed parentheses squared. And this again, we can have the vertex form. So Y equals a open parentheses, X minus H closed parentheses squared plus K, which means that we will therefore have a vertex at one comma zero. Because our age is one, we have a X minus one and our form is X minus H. So one is the H and there is no K. So we have zero for the K. So now we have a vertex for equation number two and we can move on to the same to do the same steps. So we can plot points now. So in this case, once again, you can plug any point that you want. I'm going to first plug in X equal to zero, which means that we'll have Y is equal to negative open parentheses, zero minus one closed parentheses squared, which is equal to negative open parentheses, negative one closed parentheses squared, which is equal to negative one or the 0.0 comma negative one. And that will be the first point for our second graph or second equation. And then we can have another point. Let's test X equals two will have Y equals negative open parentheses to open parentheses, two minus one closed parentheses squared, which is equal to negative open parentheses, one closed parentheses squared, which is equal to negative one or a point at two comma negative one. And that will be the second point for our graph of the second inequality. And now we can test once again, I'll test the origin to see if it falls within the region that we will be shading. So we'll have that zero is less than negative open parentheses, zero minus one closed parentheses squared or zero is less than negative, open parentheses, negative one closed parentheses squared or zero is less than negative one. And this is also false. So the origin will also not fall within the shaded region of our second in equality. Now I will be scrolling down a bit just so I can have more room to work with for our graph. So we draw a Y axis and then X axis and we can plug in the points that we know. So on the positive side of the Y axis, I'll go up to five. So I'll draw the line at 1234 and five and I'll label that and I'll do the same thing on the X axis positive and negative. So I'll do 1234 and a line at five. And on the negative X, I'll do negative one, negative two, negative three, negative four and negative five. And the same thing on the Y axis negative set will be negative one, negative, two, negative three, negative four and negative five. And now I can plug in the points that we know. So we're going to have a point or a vertex at negative four is for inequality number one, we had a vertex at negative four comma three. So we plot a point there, then we have a point at negative five, comma four and negative three comma four. And we can see that this will take the shape of a parabola. So because it's a quadratic function and as we said, uh inequality, number one is a non strict inequality, which means we will connect these points using a solid line. So we do exactly that and then extend beyond the points. And we said that we have to shade in the region that excludes the origin. So we'll be shading in the inside of this parameter. Now we move on to our second inequality and its different points. So the second inequality will have a vertex at one comma zero and it'll have a point at zero, comma negative one and another point at two comma negative one. And then we do the same thing. But before we connect them, we know that as we said earlier, this is this is a strict inequality in inequality. Number two was a strict inequality, meaning that we have to use dashed lines. So we used a, we use a dash line to connect our points and extend beyond them. And we also said for this graph that the origin will not be part of it. So we shade in the region inside of that Parabola. And we have to know that this will have no solution because they do not intersect because the shaded regions are not intersected because shaded regions don't enter set. So when you have a system with two inequalities, the answer would be the section in which the shaded regions intersect. In this case, there is no section like that, our shaded regions are opposite of each other. So for inequality, number one, we shade in the region inside of that Parabola. And for inequality number two, we shade in the region of inequalities. Number two Parabola. So they will never intersect and therefore there will be no solution for this graph. So I hope that was helpful. And until next time.

Graph the solution set of each system of inequalities.
y ≥ x² + 4x + 4
y < -x²

Key Concepts

Graphing Quadratic Inequalities

Parabola Vertex and Transformations

Solution Set of a System of Inequalities

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