Graph the solution set of each system of inequalities.
x + 2y ≤ ...

Question

Graph the solution set of each system of inequalities. x + 2y ≤ 4 y ≥ x^2 - 1

Inequalities x - 4y ≥ 2 and y ≥ x² - 3 displayed in a mathematical format.

Accepted Answer

Hello everybody. Everything I today today, we're going to look at this question that states provides a graph of the solution set of the following system where the system contains two E two inequalities. So inequity number one will be X minus four, Y is greater than, or equal to two. And inequality number two will be Y is greater than or equal to X squared minus three. Now, the first thing I'm going to point out is the fact that both of these inequalities are non strict inequalities. And this is because they have a greater than or equal to sign, if it was greater than by itself, it would be a strict inequality. And because it's non strict, we will therefore have a solid line in our graph, our graph will be solid. So that's the first thing that we have to know and then we can go ahead and move on and go back to trying to solve these inequalities. So we'll focus on the first one inequity number one first. And the first thing we'll do is just substitute the greater than or equal sign with just a simple equal sign just for the sake of solving. So we'll have X minus four, Y is equal to two. And now this can be, this reminds me of an equation for a line called Y equals M X plus B. Now I'm going to try to have our equation number one in the same form of Y equals X plus B. So the first thing I'll do is move the X over from the left hand side to the right hand side and we'll have negative four, Y is equal to negative X plus two. And then I'm going to divide both sides by negative four. And we'll be left with Y is equal to 1/ X minus one half. Now, in our equation of Y equals M X plus B, we know that the B is equal to the Y intercept N R A Y dash I N T. So that means that our Y intercept or B R B, which is the negative one half. So our Y intercept will be at zero comma negative one half. And that's our Y intercept. So we have our first point for qua inequality number one. Now, for the X intercept, there's no shortcut like the one we just used, but we can set our Y equal to zero N R I X dash I N T as an abbreviation. So we'll have zero now is equal to 1/ multiplying X minus one half. And we have to solve for X. So I'll move the one negative one half from the right hand side to the left hand side. And we'll have that one half is equal to 1/4 X. And then if we multiply the left hand, both sides by four, we'll have that two is equal to X or the 0. comma zero. Now, because there's an inequality, we know that we're going to have to shade in our region because in, in inequalities don't have one single answer. There are multiple answers that could make the inequality true. So we have to test the point to see what region we're going to shade in. So we're going to test and you can test any point you want, but we're going to test here the origin. So zero comma zero. And for that, all we have to do is plug in zero for our X and zero for our Y in inequality. Number one. So I have that zero minus four, multiplied by zero is greater than or equal to two, negative four, multiplied by zero is zero. So we have zero and zero minus 00. So we have that zero is greater than or equal to two, which is false, meaning that the origin will not be part of the shaded region in inequality. Number one, now we can move on to inequality number two. So inequality number two is a bit different in that it is not a line. So the first thing I'll do is once again substitute the greater than or equal to sign with just an equal sign. So we'll have Y equals X squared minus three. Now, there is a parent function here. An apparent function is just the function where of which our graph is based off of without any transformation. So the minus three in our inequality number two is a transformation. So if we take that transformation out, we'll see that our parent function is Y equals X squared. And what do we know about why? Equal equal X squared or in the graph of Y equals X squared? Well, we know that X squared or its graph is a parabola with a vertex at the origin or zero comma zero and has the points one comma one and negative one comma one. So every point is reflected about the Y axis. So how do we go from our parent function to our current function? Well, our current function is Y equals X squared minus three. So what does the minus three do? Well, it means that the graph of X squared shifts three units down. So that's all the minus three does we have, we still have a parameter because of the X squared? But all it does is all its values are shifted down by three units. So where we had a vertex and zero comma zero, now that point or that vertex now becomes zero, comma Well, if we started at Y of zero and we moved down three that means we're in a Y of negative three. If we had a point at one comma one, and once again, we move the Y down by three units, our X will remain the same. So we'll have Y comma negative two. And if we had a point, a negative one, comma one. And once again, we move down by three units, our X will remain the same. So we'll have negative one, comma negative two, which makes sense because once again, in a Parabola, the values are reflected on the around the Y axis. So lastly, we'll once again have to test the point to see which area we're going to shade in because we have an inequality. So we test zero comma zero once again. And all we have to do is plug in zero for the X and the Y or the of the inequality. So we'll have that zero is greater than or equal to zero squared minus three, zero, squared is zero minus three is negative three. So we have done zero is greater than or equal to negative three, which is true. Meaning that for inequality number two, the origin will be part of the shaded region. So now we're ready to graph. So we draw a Y axis and an X axis and we can start graphing little by little and focus on each inequality at a time. So we'll start with inequality number one and we'll draw the different points that we know. So on the X axis, I'll label a one to and three and I'll do the same on the negative side. So I'll have negative one, an of negative two and an of negative three. I don't need Y axis. I'll go on the positive side and I'll label and I'll do the same thing on the negative side. So I'll have negative one, negative two, negative three and negative four. So now start plugging in the different points that we know. So for inequality number one, we said that there was a Y intercept at zero comma negative one half, negative one half is below the X axis. So we go in between zero and negative one. And in the middle of that, we brought a point and we know that our X intercept is at two comma zero. So we brought another point there. And before we connect those points, remember what we said at the beginning of this question, we have non strict inequalities, meaning we have solid lines so we can connect this like any other line, we can use a solid line and we extend it to the left and to the right and it might look a look a little curved because we are doing this with our hands, but it should be a straight line. Now, what we said about the shaded region for this is that the origin was not part of the shaded region. So we have to shade in below our line. So that's exactly what we'll do and we'll shade it in lightly because the area that we really care about is the area that both of our inequalities intersect at the region that they both share. So we'll, we'll shade in this region lightly first and then we'll see where they connect. So now we can move on to inequality number two, our parameter. So we set our para having a vertex of zero comma negative three. So we put a point there and just to remind us, we had a point at zero comma negative three and I'll highlight the, the other points that we calculated. So we had another point at one comma negative two. So I'll put a point there and another point at negative one comma negative two. So I'll put a point there and once again, we know that this is a non strict inequality. So we can use a solid line and we can connect it like the normal parameter. This is our parent function, the exact same parent function just moved down by three units. So we will finish the Parabola and extend it upwards. And then we said that for this Parabola, the region that we cared about the origin was part of the shaded region. So we shaded inside of the Parabola and was shaded lightly. And when we get to that area where both er both inequalities share a region, we shade that in dark because that area is the one that we care about that area is the answer to this graph into the system. So just like that, we have been able to graph our system of inequalities where we have a and a line that share a region. So I hope that was helpful. And until next time.

Graph the solution set of each system of inequalities. x + 2y ≤ 4 y ≥ x^2 - 1

Key Concepts

Inequalities

Graphing Systems of Inequalities

Quadratic Functions

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