Graph the solution set of each system of inequalities.
y ≥ (x - ...

Question

Graph the solution set of each system of inequalities. y ≥ (x - 2)^2 + 3 y ≤ -(x - 1)^2 + 6

Accepted Answer

Hello everybody and everything. All today today, we're going to look at this question that states provide the graph of the solution set of the following system where our system includes two different inequalities. And we have inequality number one is Y is greater than o greater than, or equal to open parentheses, X minus three, closed parentheses, squared minus five. And inequality number two is Y is less than or equal to negative open parentheses, X minus four, closed parentheses squared plus three. Now, the first thing I'll note between these two inequalities is that they are both non strict inequalities. And what does this mean? Well, they are not strict because we have an or equal to sign in them. So inequity number one being greater than, or equal to and inequity number two being less than or equal to. If it was just greater than or less than, then there would be strict inequalities. But because they're not, we have non strict inequalities and therefore we will have solid line, a solid line in connecting our graph. And that will be more clear once we get to the graphing part. So now we can go back to inequalities and focus on one at a time focusing on inequality. Number one, first, I'll rewrite it as Y equals open parentheses, X minus three, closed parentheses, squared minus five. Now I write it like this or, or I, I, I set it as an equal sign just for the purpose of solving. But I see that this way of writing the, the equation reminds me of the vertex form of quadratic of a quadratic function. So now I will explain exactly what that is. So we have a quadratic, we know we're gonna have a quadratic function because we will have an X squared once that second power is distributed. So the vertex form of a quadratic function states that Y is equal to a open parentheses, X minus H closed parentheses squared plus K where H or open parentheses, H comma K closed parentheses is the vertex. So therefore, if we compare the vertex form to our formula, we'll see that the three is the age and the negative five is the can, so we'll have that our vertex, our vertex will be three comma negative five. So now we know the vertex of our first in a quantity and we know that because it's a quadratic function, it would look like a parabola once we actually get to graphing. So now we can go ahead and start plotting points. And basically this is just to see kind of what our graph is going to look like. If we have an X of a certain value what the Y would be. And such in this case, I'm going to text test X equals two. I'm gonna plug that in. So I'll have Y equals open parentheses, two minus three, close parentheses, squared minus five, which is equal to open parentheses, negative one, close parentheses, squared minus five, which is equal to negative four or the 0.2 comma negative four. And that will be the first point here. And then we can test another point. Let's say X equals four. Now you can try any point that you want, you can plug in any point that you want. These are just the ones I picked. So let's say if X equals four will have Y is equal to open parentheses four minus three, closed parentheses squared minus five. And that will be equal to four minus three is one squared. So we'll have one squared minus 51 squared is one. So we'll have one minus five which is negative four, show a point at four comma negative four. And that'll be our second point for inequality number one. Now, because we have an inequality, we know that there will be a specific region of our graph which we will have to shade it. And that's because an inequality doesn't have just one answer. Let's see, for example, in inequality number one, we have that Y is greater than, or equal to something else. So anything that the Y is greater than or equal to will be an answer. So we have to shade in that region that will function as the answer. So in this case, you have to test a point and see if it falls within that region. In my case, I'm going to test the origin zero comma zero, but you can test any, any point that you want on the graph. So what I'll do is just plug in X uh or plug in zero for our X and zero for our Y. So we have that zero is greater than, or equal to open parentheses, zero minus three, closed parentheses squared minus five, which means that zero is greater than, or equal to nine minus five because zero minus three is negative three and negative three squared is nine, which will leave us with zero is greater than or equal to four, which is false. So we know that the origin will not fall within the region that will be shaded for this graph. So that will be it for uh inequality number one. And now we can move on to inequality number two and we'll go in the same order. So first we'll change the less than or equal sign with an equal sign and we'll have that Y equals negative open parentheses, X minus four, close parentheses squared plus three. And remember that this matches the vertex form of a quadratic function. So we have Y equals a open parentheses X minus H closed parentheses squared plus K. And now we can look at our vertex here. So our vertex will be four comma three because the four is the H and the plus three is the K. So now we've obtained our vertex four inequality number two. And now going in the same order, we can go ahead and plot some points. So first, I'll test X equals three, which means we'll have Y equals negative open parentheses. Three minus four, closed parentheses squared plus three, which is equal to negative open parentheses, negative one, closed parentheses squared plus three, which means that Y is equal to negative one plus three, which is equal to two order the 0.3 comma two. And now we can see that we have obtained our first point other than the vertex for this graph and we can go ahead and test another point. Let's say at X equals five, we'll have that Y is equal to negative open parentheses, five minus four, closed parentheses squared plus three as equal, equal to negative open parentheses, one closed parentheses squared plus three, which means that Y is equal to negative one plus three, which is equal to two or the 0.5 comma two. So that will be our second point for this graph. And now we can go ahead and do the final step as we did with inequality number one, which is test a point to see if it falls within our shaded region. Once again, I will be testing zero comma zero or the origin, but you can test any point that you want. So we'll have zero is less than or equal to negative open parentheses, zero minus four, closed parentheses squared plus three, which means that zero is less than or equal to negative open parentheses, negative four plus parentheses squared plus three or zero is less than or equal to negative four squared is positive 16. And we have the multiplied by a negative would be negative 16, negative 16 plus three is negative 13. So this would be false as well. So we know that for this inequality, the origin will not be within the region that will be, that will be shaded. So now with all this information about inequality, number one and number two, we can go ahead and make a graph. So we draw a Y axis and an X axis and on the Y axis, I'll draw a dash at 1234 and five. As well as on the negative side, we'll say negative one, negative two, negative three, negative four and negative five. And on the X axis, let's go up to around seven. So we'll draw on the positive side, we draw a dash at and we'll put draw another one at eight. So now we can go ahead and start graphing our inequalities. So let's focus on inequality. Number one first we said that we had a vertex at three comma negative five and a point at two, comma negative four and four comma negative four. Now we know that this is going to look like a traveler. And as we had mentioned, these are not strict inequalities. So we will use a solid line so we can extend past both points. And that would be the parameter that we get. Now, we also said that the origin will not be part of the region shaded. So we shade within the Parabola and are shaded very lightly at first to see if anything overlaps with our second inequality. So for our second inequality, we said that we had a vertex at four comma three. So we put a point there a point at three comma two and five comma two. And once again, we use a solid line for this inequality as well because it's a non strict inequality and we connect the dots to look like a Parabola and we extend past both points. And we also said that for this inequality, we do not include the origin in the region that we will shade. So we shade inside of the problem. And now we see the region that will overlap and we shade that region in much darker. To emphasize that that region of inter lapping will be the correct answer. So the solution for this set of inequalities is the region in which they intersect in which the shaded regions intersect. So that's why we want to put an emphasis on that region. So we make sure we shade it in nice and dark. And just like that, we were able to graph both of our inequalities and shade in the region that they share in common to emphasize are answered. So I hope that was helpful. And until next time.

My Courses

Chemistry

Biology

Math

Physics

Business

Social Sciences

Programming

Product & Marketing

Graph the solution set of each system of inequalities. y ≥ (x - 2)^2 + 3 y ≤ -(x - 1)^2 + 6