 Let us summarize what we've learned so far about quadratic functions in the graphs. The vertex of a quadratic function can be found by the formula h equals negative b over 2a. And then the associated y-coordinate k can be found just by looking at f of h right there. And we can use this to find the quadratic, or the vertex each and every time. If we want to find the x-intercepts of a quadratic function, you know, it looks like f of x equals a x squared plus b x plus c. We can maybe, we have to solve the quadratic equation, f of x equals zero, which we can solve that by factoring to complete the square. But you can also use the quadratic formula x equals negative b plus or minus the square root of b squared minus 4ac all over 2a. And using the discriminant inside of the quadratic formula right there, we can determine whether the graph's gonna have two, one, or zero x-intercepts. And the last thing I wanna mention is we can also talk about the concavity of the graph here. When your leading coefficient is a positive number, this means the graph will concave upward and the vertex will represent a minimum on the graph. When your leading coefficient is negative, this will actually tell your graph is concave downward, and thus the vertex will be a maximum on the graph. Putting these three bits of information together can nearly always give you a pretty accurate picture of a quadratic graph that is a parabola. So imagine we have the function f of x equals negative 3x squared plus 6x plus one. If I was to graph this thing, there's a couple of things I could do. At first look for some x-intercepts, right? We can find that from the quadratic formula. We could try to factor although this one doesn't factor very nicely at all. By the quadratic formula, we get negative b, which is gonna be a negative six right here. I mean, after all, our values are a equals negative three, b equals six, and c equals one. So we're gonna get a negative b plus or minus the square root of b squared, which will be a 36. Then you're gonna get negative a, negative four times a times c. a times c is just negative three times by negative four gives you a positive 12. Three plus 12 is a 48. Now 48 itself is not a perfect square, but it does have 16 as a divisor. 48 is 16 times three. The square root of 16 is then four. So you get negative six plus or minus four times the square root of three over negative six. If you factor out a negative two from the numerator, that would leave behind a three plus or minus two times the square root of three all over negative six. And then negative two goes into negative six three times, thus giving us what we have right here. And as we wanna graph these, these are our x-intercepts. We're going to need to approximate these things. And so if you look at the two choices, minusing the two root three will give us negative 0.15 if we round to two decimal places. And if we take three plus two root three over three, I'll give you 2.15. So this gives us two points on the graph, two x-intercepts. We get 2.15 comma zero. And then the other one over here is negative 0.150. The y-coordinate of an x-intercept is always zero by construction. Then if we could find the vertex, we could graph this thing pretty well as you can see on the screen right here. Now the vertex, remember, we just find that to be h is negative b over two a, which in this situation, we get negative six over negative six, which is equal to one. And then once we have one, we can plug that into the formula K equals negative three plus six plus one. And we see that the vertex is gonna be one comma four. One comma four, that's given right here. Notice that this function will always have as its axis of symmetry, the vertical line that goes through the vertex. So this gives you the line x equals one. And then using these three points, the x-intercepts and the vertex, you could probably piece together a pretty good looking parabola. You don't do your best to make that bowl shape. If you're drawing by hand, you won't be screwed nice too much probably, but we get something like you see drawn by the computer right here. So with the vertex and the x-intercepts, we can graph any parabola for the most part. So there are a few exceptions, right? This, what we saw previously is if we have a parabola, which has two distinct real x-intercepts. What if we don't? Like if you take f of x equals x squared minus six x plus nine, in that situation right here, this actually is a perfect square trinomial. It factors as f of x equals x minus three squared. And so the x-intercept, there's only one of them, actually coincides with the vertex. So you see the vertex is three comma zero in this situation. You need another point to determine what's going on here. Now we could use the leading coefficient to help us out a little bit, right? A equals one, which is positive. This tells us the graph concaves up. So the picture has to be something above the x-axis. That is useful. The next thing also to mention is that since you have a equals one, we can then go up one, one y-coordinate over one, x-coordinate to find another point, which is gonna be four comma one. And then using symmetry, because the function is symmetric with respect to the line x equals three, we can find another point using reflection here. Say at two comma one. And then using those three points, we can connect and make our parabola, right? Like so. Now you don't have to use necessarily the next point over. You can use any point in the domain, right? You could just be like, oh, what is f of five? You could do something like that. And you can compute that, that give you a point over here and then you reflect to get you another point over here. It doesn't matter. You just have to make sure you pick a different point other than the vertex. And then you can work from there. It doesn't matter as long as we find these points, you're gonna be good to go in that regard. Now I also, that's if your vertex is on the x-axis, what if your vertex is even on the x-axis? What if it's, what if there's no real solutions, no real x-intercepts? And so your graph might look something like this. Well, in that situation, I'd still tell you to find the vertex, right? H comma k. And then pick another point, like maybe use the slope or just plug and chug to find some other point over here, h plus one comma k plus whatever the a value is, something like that. I think that would be a good thing to do. And then use symmetry to find the other side. We can use this to graph any quadratic function.