Graph the solution set of each system of inequalities.

Question

$y>x^2+2$

$y\le x-2$

Accepted Answer

Welcome back everyone. In this problem, we want to graph the solution set of Y is greater than X squared plus four and Y is less than, or equal to negative X minus six. Here we have graphs of possible solutions or A B. See. And D says it's another set. OK. D says that there's no graph rather. And for our solution here, I've just attached a graph that we can use to make our own. So we'll be able to use this to plot our own graph and then compare it to our answer choices. Now, remember, let me just rewrite here. We know that we are trying to find our graphs or Y is greater than X squared plus four and Y is less than, or equal to negative X minus six. OK? No, first let's try to start by graphing Y is greater than X squared plus four. Now, in this case, let's let Y OK be equal to X squared plus four. OK. In this case, this can help us to plot our line that we will need to help us figure out where the solution set will be. And because we have a squared term here, we know that it's going to graph a Parabola. Now let's try to find some points by substituting values of X. Let's say that X equals zero. Now, when X equals zero, that means Y will be zero squared plus 4, 0 squared is zero. So that means Y equals four. So therefore, +04 is going to be a point on our graph. So I can go ahead here and put +04. Next, let's try to find it when X equals one. When X is one, Y is one squared plus 4, 1 squared is +11 plus four is five. Therefore, that means +15 is on our graph. So I can also put that on our graph here. And let's go to the other side of our Y axis. OK, where X is negative one and no at X equals negative one, Y equals negative one squared plus four, negative one squared is one and one plus four is five. So this tells us that negative +15 is on our graph as well. Now, in this case, we can tell that negative 15 and 15 are symmetric about the Y axis. So we have our axis of symmetry, we have three points, we can go ahead and graph our parabola. So now when we draw our line here and remember it doesn't have to be perfect. We just want to have an idea of what it will look like. So we can compare it to figure out our solution set. OK. So when we do our graph, OK? And because it's greater than we know that it doesn't include these points. So we can use a dotted line to show that. OK, then this is the line of Y equals X squared plus four. And for the points where Y is greater than X squared plus four, it would be in this shaded region. And we know we're right because if we were to take one of those points at random, let's say the 0.05 since the Y value is five and the X value is zero is five greater than zero squared plus four. Yes, it is because five is greater than four. So we know that we've shared the correct region now that we've done one, let's try to figure out the other when Y is less than, or equal to negative X minus six again to solve. Let's let Y be equal to negative X minus six. And no, we know that this is a linear equation. It will form a line. So we just need two points that we can connect to help us plot our graph. Let's try to first find our point when X equals zero. Now at X equals zero, that means Y equals negative zero minus six, which is going to be negative six. So that tells us that zero, negative six is going to be on our graph. So let's plot that point, OK, zero, negative six. Next we can solve for it when X equals sorry, when Y equals zero actually, OK? Because no, when Y equals zero, that means zero equals negative X minus six. And if we add X to both sides, that means X equals negative six. So when X is negative six, Y is zero, and again, we can go ahead and plot that on our graph. OK? So this is where the point negative 60 is going to be know that we have our two points. OK. Then we can go ahead and try to plot our graph. OK? Because no notice that here. OK. Then again, it doesn't have to be a perfect line for no, but I can say our straight line is probably gonna look something like this. OK? And we know it will be a solid line because it says why is less than or equal to? So it's, it includes those points on the line. Now, which region will we share? Well here notice that it says why is less than or equal to that expression? So it will be the point below our graph. OK? Or below our line rather. So it's going to be this shaded region. OK. Hope I'm doing a great job of shading here. I know we know that this shaded region is correct. OK? Because if we were to take a point here, let's say we take the point negative 55 OK at negative five, sorry, negative five, negative five at negative five, negative five is our Y value negative five less than or equal to the negative value of our X value negative five minus six. Well, negative negative five is positive five and positive five minus six equals negative one. So we will be left with negative five less than or equal to negative one. And that is a true statement because negative five is less than negative one. So from our test, we know that we've shared the correct region. So now this is what our graph of Y is greater than X squared plus four. And why is less than or equal to negative X minus six would look like now remember we're supposed to graph the solution set and the solution set exists where are shaded regions overlap. Now, in this case does or do we have any section where the shaded region overlaps? No. So since the two graphs do not have an intersection, there is no solution to this system. Therefore, D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Graph the solution set of each system of inequalities.
$y>x^2+2$
$y\le x-2$

Key Concepts

Inequalities

Graphing Quadratic Functions

Shading Regions in Graphs

Watch next