 So here's another problem that shows up frequently in calculus that it's not actually a calculus problem, and that is this challenge of finding the intersection point of two curves. So if you already know how to find the intersection point of two curves, don't worry about the rest of this. And if you're a little rusty on the methods of doing so, you might want to take a look at the following. So let's find the intersection points of the two curves y equals 12 minus x squared and y equals 4x. Just a quick comment on mathematical terminology. Mathematicians call anything that you graph a curve, even though it might or might not have some sort of English language idea curvature to it. So in this case, y equals 4x, you should recognize as being something that will give you the graph of a straight line to a mathematician that still counts as a curve. So let's take a look at the intersection point or points. Now a very important idea to keep in mind is that any point on the graph of a curve has to satisfy the equation for the curve. So if a point is on the graph y equals 12 minus x squared, whatever the x and y coordinates are, have to make the statement y equals 12 minus x squared a true statement. And if that point is also on the graph of y equals 4x, the x and y coordinates have to make that statement also true. So what does this tell us? Well an intersection point is a point that is on the curve y equals 12 minus x squared and also on the curve y equals 4x. Well that intersection points x and y coordinates have to make this a true statement and they also have to make this a true statement. So that means whatever our intersection point is, it has to correspond to a solution to the system of equations y equals 12 minus x squared y equals 4x. Which means that I can find that intersection point by solving this system. Now here's a very useful thing to notice. Intersection point is a geometric concept. It is a thing from geometry. However we're going to find that using an algebraic method solving a system of equations. And so here again is one of those neat interactions between algebra and geometry that make up modern mathematics. In fact in the course of this problem, except for phrasing the problem in geometric terms, we're not going to draw any pictures. The solution to where that intersection point is going to occur is going to take place entirely within the algebra. Well in algebra you learned a number of ways of solving a system of equations. But in this particular case, because our equations are already solved for y, I might as well make use of substitution. So again the key thing to remember, as soon as you write down that equal symbol, you are guaranteeing that the right and left hand sides of that equals are completely and totally interchangeable anywhere that you find them, at least within the context of the problem you're working. So because I've written y equals 12 minus x squared y equals 4x, I can replace any y with 12 minus x squared. Well here's a y, I'll replace that with 12 minus x squared and I get this equation 12 minus x squared equals 4x. And that's a quadratic equation. Well I know how to solve quadratic equations. So I do have to get the quadratic equation into the form ax squared plus bx plus c equals 0, so I'll do a little bit of algebra to do that. And notice I don't really care that this coefficient of x squared is negative. The quadratic formula doesn't care what the value of a, b, or c is as long as a is not equal to 0. So I apply the quadratic formula and I get the following solution and after all the dust settles I get the solution x equals negative 6 or 2. And at this point I throw a box around my answer and say I'm done with the problem and I get it back and I find out that I've gotten only partial credit. And the reason for that is we're actually looking for the intersection points. Again a point is a geometric concept. In terms of algebra a point requires a specification of both an x-coordinate and a y-coordinate. So this solution x equals negative 6, well that's an x-coordinate, but it's only half of a point. I need to find the other half. So I have my x-coordinate. If only there is something that would tell me what my y-coordinate was when I knew what the x-coordinate was. Then I could find the y-coordinate and that would give me the complete point. Oh, well here I actually have two ways of finding my y-coordinate given what my value of x is. Now because the intersection point is going to be on both curves it actually doesn't matter which of these two formulas we use to find our y-coordinate. So let's pick the more complicated one because we like doing lots of extra work. Well maybe not. Let's use the easier one, y equals 4x. So if x is equal to negative 6, I'll just substitute that into my formula, y equals 4x, and I find y is negative 24. So my intersection point, x-coordinate, y-coordinate, negative 6, negative 24. And likewise for my other solution, if x equals 2, I'll substitute that in. And my other point of intersection, if x is equal to 2, y is equal to 4 times 2, y is equal to 8. So the other point of intersection is going to be 2, 8. Now just as a final note, because the intersection point should be on both curves, one thing you may want to check is with the other formula, make sure that these two points are actually points on the graph of y equals 12 minus x squared. So here if x equals negative 6, 12 minus x squared is in fact negative 24. Likewise if x is equal to 2, 12 minus 2 squared is in fact equal to 8. So both of these points are on both curves.