# Video: Graphing Systems of Inequalities with Parabolas

Determine which region in the graph contains all the solutions to the following system of inequalities: 𝑦 ≤ (1/2)𝑥² − 𝑥 + 1, 𝑦 ≤ −𝑥² + 2𝑥 + 4.

03:14

### Video Transcript

Determine which region in the graph contains all the solutions to the following system of inequalities. 𝑦 is less than or equal to one-half 𝑥 squared minus 𝑥 plus one, and 𝑦 is less than or equal to negative 𝑥 squared plus two 𝑥 plus four.

In this question, we’ve been given two quadratic inequalities. These are represented on our graph somehow. Let’s simply begin by identifying which curve represents which equation. Let’s take the equation 𝑦 equals one-half 𝑥 squared minus 𝑥 plus one. That must be represented by this first slightly higher curve. And we know that because the leading coefficient, the coefficient of 𝑥 squared, is positive. Whereas the coefficient of 𝑥 squared in our second equation, let’s call that 𝑦 equals negative 𝑥 squared plus two 𝑥 plus four, is negative. So, we have this sort of inverted parabola as shown.

Now, in fact, the easiest way to decide which region in the graph contains all the solution to our system of inequalities is to pick a point in each region and test it out. Now, I always like to choose the most straightforward point on the graph. If possible, that’s the origin. That’s the point with coordinates zero, zero. What we’re going to do is substitute 𝑥 equals zero and 𝑦 equals zero into each inequality and see if these points satisfy them.

So, taking 𝑦 equals zero and 𝑥 equals zero in our first equation, we get zero is less than or equal to negative a half times zero squared minus zero plus one. Zero is less than or equal to one. Well, zero is less than one. So, this is true. Remember, the equals bit just tells us what kind of line we have. Here, we have a solid line, so we know it’s going to be equals rather than just less than.

When we substitute these values into our second inequality, we get zero is less than or equal to negative zero squared plus two times zero plus four. Which simplifies to zero is less than or equal to four. Well, yes, zero is less than four, so this inequality is also satisfied. And so, it looks like region 𝐸 is the region which contains all the solutions to our system of inequalities.

It’s sensible for us to at least check a couple of the other regions. Let’s check a point in region 𝐴. Let’s take the point negative two, zero. Substituting 𝑥 equals negative two and 𝑦 equals zero into our first inequality, and we get zero is less than or equal to a half times negative two squared minus negative two plus one. Well, that tells us that zero is less than or equal to seven, which is, of course, true.

So, we substitute these values into our second inequality. This time we get zero is less than or equal to negative negative two squared plus two times negative two plus four. The right-hand side simplifies to negative four. So, we see that zero’s less than or equal to negative four. Of course, that’s not true. Zero is, in fact, greater than negative four. And so, we can obviously disregard region 𝐴. Whilst it contains solutions to the first inequality, it doesn’t contain all solutions to the entire system.

Let’s repeat this process one more time. This time we’ll choose a point in 𝐵. Again, we’ll choose a point on one of our axes because it makes the numbers a little bit easier. And we’ll choose a point with coordinates zero, six. Substituting 𝑥 equals zero and 𝑦 equals six into our first inequality, and we get six is less than or equal to one. That’s obviously not true.

Similarly, substituting into our second inequality, we get six is less than or equal to four, also not true. So, once again we can disregard 𝐵. And in fact, if we continued, we’d see that we can also disregard 𝐶 and 𝐷. And so, the region which contains all the solutions to our system of inequalities is indeed 𝐸.