### 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 weโre going to have a negative leading coefficient. The coefficient of ๐ฅ 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 ๐ฆ equals ๐๐ฅ squared plus ๐๐ฅ plus ๐. We could then choose some coordinates on each curve and substitute them in to form a system of simultaneous equations.

Here, thatโs quite long winded. So instead, weโre going to use something thatโs sometimes called the vertex method. We can write a parabola in vertex form as ๐ฆ equals ๐ times ๐ฅ minus โ all squared plus ๐. The vertex of this parabola lies at the point with coordinates โ, ๐. So, letโs 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. Weโre going to then let โ be equal to negative one and ๐ be equal to zero. And substituting this into our formula, well we get ๐ฆ equals ๐ times ๐ฅ minus negative one all squared plus zero. Well, ๐ฅ minus negative one is ๐ฅ plus one, and our equation becomes ๐ฆ equals ๐ times ๐ฅ plus one all squared.

Our next job is to find another point that this curve passes through. We can choose any really. Letโs choose this point. It has coordinates negative three, negative two. By substituting ๐ฆ equals negative two and ๐ฅ equals negative three into this equation, we form an equation purely in terms of ๐. So, we get negative two equals ๐ times negative three plus one all squared. We get negative two equals ๐ times negative two squared. And, of course, negative two squared is four, so we have negative two equals four ๐. And we divide through by four, and we get ๐ equals negative two over four or negative one-half. Letโs substitute this back into the earlier equation. And so, we find the equation of our first curve to be ๐ฆ equals negative one-half times ๐ฅ plus one all squared. Weโll repeat this process for the second curve. Thatโs in red on this diagram. Itโs the slightly higher curve.

This time the vertex, the turning point, lies at one, one. So, weโre going to let โ be equal to one and ๐ be equal to one. And our equation becomes ๐ฆ equals ๐ times ๐ฅ minus one all squared plus one. We choose a second point on the curve. Now, this curve passes through the origin, zero, zero. So, letโs choose that one. And we get zero equals ๐ times zero minus one squared plus one. Thatโs zero equals ๐ times negative one squared plus one. And, of course, negative one squared is simply one. So, we get zero equals ๐ plus one. If we subtract one from both sides, weโve solved for ๐. ๐ is equal to negative one. We substitute this back into the earlier formula. And we find the equation of our second curve to be ๐ฆ equals negative one times ๐ฅ minus one all squared plus one, which actually weโre just going to write as negative ๐ฅ minus one all squared plus one.

And so, we have the equations of the curve, but weโre looking to find the system of inequalities that forms the shaded area shown on the graph. To answer this, weโre going to pick a point that satisfies both of our inequalities. So, weโre going to pick a point within the shaded region. Letโs choose the point zero, negative two. What weโre going to do is substitute ๐ฆ equals negative two and ๐ฅ equals zero into each of our equations. And weโre going to see which side of the equation is greater. In our first equation, we get negative two something, weโre 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 ๐ฆ must be less than or equal to negative one-half times ๐ฅ plus one squared. Letโs repeat this process for our second curve. This time we get negative two something negative zero minus one squared plus one. Thatโs 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 ๐ฆ is less than or equal to negative ๐ฅ minus one all squared plus one. And so, we found the system of inequalities. They are ๐ฆ is less than or equal to negative one-half times ๐ฅ plus one squared and ๐ฆ is less than or equal to negative ๐ฅ minus one all squared plus one.