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## 9-12.A.REI.11 Transcript

This is Common Core State Standards support video for Grades 9 through 12. The standard is A.REI.11. The standard states: explain why the x coordinates of the points where the graphs of the equations y equals f of x and y is equal to g of x intersect are the solutions of the equation f of x equals g of x; find the solutions approximately, example, using technology to graph the functions, make tables of values, or find successive approximations; include cases were f of x and/or g of x are linear, polynomial, rational, absolute value, exponential, and logarithmic functions.

That’s a whole lot of different types of functions, but we’ll do an example of several of these. The most basic will be your linear equations. So, if we were to take f of x to be minus 4x minus 7, and g of x to be 2x plus 11, then what we have here is, again, two linear equations, and if we look at this standard, it says to explain why the x coordinates of the points where the graphs of the equations of those two functions intersect, are the solutions of the equation f of x equals g of x.

What we have is a situation where the solution for this equation as opposed to the solution for that equation, the point of intersection is, in fact, the solution for both. Now, since it is, in fact, the same point, then it stands to reason that they will both have the same x intercept. One way to solve this would be to just simply do what the standard says do, to set f of x and g of x equal to each other. And so, then, by doing this, in this situation it’s minus 4x minus 7 is equal to 2x plus 11.

Now, using basic algebra, it might be easier to add 4x to both sides. So, we would have 6x, and then subtract 11 from both sides. So, that would be negative 18. So, we have minus 18 equal to 6x. So, our x-coordinate would be a minus 3. And so, that does correspond and make sense, matches up with our graph. Now, it’s just a matter of taking our minus 3x value and substituting back into either one of the two functions. If we substitute minus 3 into g of x, that will give us a minus 6 plus 11, which would be a 5. So, then, the solution should be the point 3, 5.

What can do is use technology to verify what we just did. So, here’s our two functions, f of x is minus 4x minus 7, and g of x is 2x plus 1, and if we graph those, there they are. Now, there’s a couple of ways to do this. By using successive approximations on the calculator, that amounts to using our trace button. So, if we were to trace our values, okay, here we go, here we go. Went too far, and that looks like it might be the point of intersection. If I jump from one function to the other, I still remain at 5, so it is the correct solution.

Another way would be like the standard asks for, would be to use a table. And we can use technology to accomplish this also, although you might have the students do it manually also. But, we can also do a table. Let’s see, here, looking at the x values, we have none of the y values for the two different functions the same. But there we have minus 3 where both y values are 5. So that is, in fact, our point of intersection. It is a common solution to both.

Now let’s use a couple of other different types of functions. In this case, let’s see, we’ll use a polynomial function. We’ll use a quadratic and a cubic equation here, function. If we graph the first one, just roughly sketching it out, it would look something like this. And then we graph function g, it would look something like this. Now, this is going to be more difficult because if we set the two equal to each other, then what we will have, when we set the two functions equal to each other, it’s going to start to be a lot more difficult for students because now, we have .25x to the third power plus 5 equal to our function g, which would be minus x squared minus 2x, plus 15. Now, this is solvable. It would be pretty difficult. We’re going to have to do a bunch of rearranging and factoring and so forth. So, it does start to be a difficult task when you start getting into the more complicated functions.

So, again, this is where our technology comes in to give students an alternative, especially for those students who are still in Algebra, Algebra 1, Algebra 2. Once you start getting into Precalculus and Calculus, students will have other methods. But, depending again, on the level of the student, we would still need to utilize technology to help us here. Now, using technology, on my calculator, I have the third and the fourth functions (would be the ones that we’re dealing with now). So, if I were to graph those two, okay, they kind of show that our initial sketch was pretty accurate.

But now, again, we’ve got two different routes to go, successive approximations. Again, we would trace. And, let’s see, there we are on one function. If we trace over, for that point of intersection, it looks like it’s around 2. When I jump to the other function, it doesn’t change the y value. So, it looks like our point of intersection is 2, 7. So, this should be the common solution to both. Let’s look at a table, and let’s use our calculator to help us with that. And if we look right here at x equal 2, we have the same y value for both. So, using our technology, we’re able to find a solution for both to be the point 2, 7.

Let’s continue on. The standard asks for other types of functions, so let’s try a rational function and an absolute function together. So, if we graph the rational function, it would look something like this. And here, we would have a vertical asymptote, and it would be at x equal to 3 because that would give us a 0 in the denominator to kind of again, make sense of the generalities, the general nature of this graph. Then, if we graph the absolute value function, it would look something like this.

So, it looks like we might have two points of intersection. So, now, we again use our technology to help us. Now, this one would not be as difficult as some of the others in terms of setting the two equations equal to each other and solving it manually. So, again, something like this, the students could handle, setting f of x equal to g of x here. Again, students in Algebra 1, Algebra 2 could handle this type of problem without the technology. But, of course, we want to utilize the technology to the extent that we can.

Here we have our next pair of functions. On this calculator, I have it as functions 5 and 6. So, if we graph these two, so there was our first one, the rational function. Then we have our absolute value function. Again, if we decide to try tracing, we are on the rational function. It looks like we have a possible intersection here at 1, 0. I jump from one to the other and, yes, that looks like a value of . . . that looks like a point of intersection there that would solve both equations, 1,0. And by substituting, we can prove that to be the case.

Then, if we continue to trace, it looks like we have another point of intersection here. Let’s try 7, 6. Jump from one to the other and it doesn’t change. So, yes, it looks like this is a second solution, the point 7,6. If we try the table like the standard suggests, if we look at 1, 0, it’s common to both. And if we kept going, there’s where x is 7, and we do have the values of both functions being 6 for your y value. So, again, using technology to help us out, we got the solution for this pairing.

Let’s try one last pairing, and this time, let’s go with an exponential function and a logarithmic function. So, if we graph the exponential function, it would look something like this. Then, when we graph the logarithmic function, it will look something like that. Again, it looks like we might have a possible two points of intersection, so, let’s use our technology to verify. Using our technology, a graphic calculator in this case, we have functions seven and eight on the calculator already set up for our logarithmic and our exponential functions.

So, let’s graph these two, and there they are. So, we can try to trace, and again, it looks like somewhere in here there might be a point of intersection. Now, it’s very difficult, even if we were to zoom this in, to see where the point of intersection is. Now, depending on your calculator, some of them do have the capability to actually find the point of intersection. Earlier, we were able to use successive approximations, and we were able to find some exact solutions in terms of being whole numbers or integers. But, in this case, we’re not going to be that lucky.

So, what we can do is, for this calculator, notice that it can calculate the point of intersection. So, in this case, okay for the first curve, the second curve, and it says to guess, but it will find something very close. So, there’s one point of intersection, and again, this is a very close approximation. It would not be exact. Now, looks like there’s another point of intersection here. It looks like there’s another point of intersection here. So, let’s use the power of the technology to again, get an approximate value of where they intersect; first curve, second curve, and there is our approximate solution. Depending on how accurate or, depending on the accuracy, again, we could round this off to several values. But, again, here is the approximate value for the point of intersection. The solution for both of these functions would be about 5.82 for your x value, and 3.53 for your y value.

So, looking back, again, using technology, there are quite a few options. And again, it depends on the graphing calculators and so forth that you have available to you. But, again, you can graph the functions. You have tables of values. You can use successive approximations by tracing, and some calculators even have the capability to calculate, at least, very closely, your points of intersections, which would be the solutions to the systems of equations that you’re dealing with.