# Video: Writing a System of Inequalities That Describes a Region in a Graph

Find the system of inequalities that forms the shaded area shown in the graph.

04:16

### Video Transcript

Find the system of inequalities that forms the shaded area shown in the graph.

Before we can think about finding a system of inequalities here, we simply need to find the equations of the two curves. That distinctive parabola shape tells us that these are going to be quadratic equations. And actually, we know we’re going to have a negative leading coefficient. The coefficient of 𝑥 squared is going to be negative. That gives us this inverted parabola. And there are a number of ways that we can find the equations of each curve. We could use the fact that the general form for the equation of a quadratic graph is 𝑦 equals 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. We could then choose some coordinates on each curve and substitute them in to form a system of simultaneous equations.

Here, that’s quite long winded. So instead, we’re going to use something that’s sometimes called the vertex method. We can write a parabola in vertex form as 𝑦 equals 𝑎 times 𝑥 minus ℎ all squared plus 𝑘. The vertex of this parabola lies at the point with coordinates ℎ, 𝑘. So, let’s see how this helps us find the equation of the blue, the first slightly lower, curve. Its vertex or turning point lies at the point negative one, zero. We’re going to then let ℎ be equal to negative one and 𝑘 be equal to zero. And substituting this into our formula, well we get 𝑦 equals 𝑎 times 𝑥 minus negative one all squared plus zero. Well, 𝑥 minus negative one is 𝑥 plus one, and our equation becomes 𝑦 equals 𝑎 times 𝑥 plus one all squared.

Our next job is to find another point that this curve passes through. We can choose any really. Let’s choose this point. It has coordinates negative three, negative two. By substituting 𝑦 equals negative two and 𝑥 equals negative three into this equation, we form an equation purely in terms of 𝑎. So, we get negative two equals 𝑎 times negative three plus one all squared. We get negative two equals 𝑎 times negative two squared. And, of course, negative two squared is four, so we have negative two equals four 𝑎. And we divide through by four, and we get 𝑎 equals negative two over four or negative one-half. Let’s substitute this back into the earlier equation. And so, we find the equation of our first curve to be 𝑦 equals negative one-half times 𝑥 plus one all squared. We’ll repeat this process for the second curve. That’s in red on this diagram. It’s the slightly higher curve.

This time the vertex, the turning point, lies at one, one. So, we’re going to let ℎ be equal to one and 𝑘 be equal to one. And our equation becomes 𝑦 equals 𝑎 times 𝑥 minus one all squared plus one. We choose a second point on the curve. Now, this curve passes through the origin, zero, zero. So, let’s choose that one. And we get zero equals 𝑎 times zero minus one squared plus one. That’s zero equals 𝑎 times negative one squared plus one. And, of course, negative one squared is simply one. So, we get zero equals 𝑎 plus one. If we subtract one from both sides, we’ve solved for 𝑎. 𝑎 is equal to negative one. We substitute this back into the earlier formula. And we find the equation of our second curve to be 𝑦 equals negative one times 𝑥 minus one all squared plus one, which actually we’re just going to write as negative 𝑥 minus one all squared plus one.

And so, we have the equations of the curve, but we’re looking to find the system of inequalities that forms the shaded area shown on the graph. To answer this, we’re going to pick a point that satisfies both of our inequalities. So, we’re going to pick a point within the shaded region. Let’s choose the point zero, negative two. What we’re going to do is substitute 𝑦 equals negative two and 𝑥 equals zero into each of our equations. And we’re going to see which side of the equation is greater. In our first equation, we get negative two something, we’re going to choose greater than or less than, negative one-half zero plus one squared. Well, this right-hand side becomes negative one-half.

Now, we know that negative two is less than negative one-half. And, of course, we have solid lines. So, we want to say that 𝑦 must be less than or equal to negative one-half times 𝑥 plus one squared. Let’s repeat this process for our second curve. This time we get negative two something negative zero minus one squared plus one. That’s negative two something zero. But of course, we know negative two is less than zero. Since, of course, the curve is indicated by a solid line, we then say that 𝑦 is less than or equal to negative 𝑥 minus one all squared plus one. And so, we found the system of inequalities. They are 𝑦 is less than or equal to negative one-half times 𝑥 plus one squared and 𝑦 is less than or equal to negative 𝑥 minus one all squared plus one.