 In this video, we provide the solution to question number 15 for practice exam 2 for math 1050. We're given a quadratic function f of x equals 3x squared plus 2x minus 2. We have to first convert it into vertex form. Once we have it in vertex form, we're then going to graph it over here, make sure that we label the vertex, the y-intercept, and the x-intercepts of the graph. So that's what we're going to be doing like so, alright? So the first thing we've got to do is convert it into vertex form. We can complete the square. There were some other methods we talked about, but that's just how we're going to do it right here. So to put it in vertex form, we're going to separate the x's from the constant. So we end up with this 3x squared plus 2x minus 2. You notice a gap here as I'm waiting for that guest of honor to arrive. We're going to factor out the 3. So we end up with 3 times x squared plus 2 thirds x and then minus 2 like so. So then we have to take half of the 2 thirds right here, which is going to be a 1 third. And then we have to square that thing that gives us a 1 ninth. And so then the guest of honor is going to be plus 1 ninth. But if we add the 1 ninth, we have to also steal away the 1 ninth so things are balanced. So we're going to subtract 3 times 1 ninth right here. Where do the 3 come from? That's because this 3 distributes in all the pieces. So we really are subtracting 3 times 1 ninth right there. So now because we've completed the square, the inside expression right here then factors as a perfect square. So we get f of x equals 3 times x plus 1 third squared. Where do this come from? We take a square root of 1 ninth, which is a third, excuse me. And then the sign here will be identical there as well. Then we have, so 3 goes into of course 1 ninth. That's going to give you right here. This is negative 1 third. So I'm going to rewrite this negative 2 as in fact as a negative 6 third. So we combine them together. So we end up with a negative 7 thirds like so. So this is now the vertex form of the quadratic equation. So we should make some indication that we've discovered that, all right? Once it's in vertex form, we can start recognizing transformations very quickly. We can also find the vertex itself. So let's make mention of that. The vertex h comma k, this is going to equal this number right here. You have to switch the sign though. So we end up with a negative 1 third. And then we can then get negative 7 thirds over here like so. And so as you graph this thing, you could make the scale honestly be basically really whatever you want. If you want this to be 1, 2, 3, 4, that's fine. If you want this to be like 1, 3rd, 2, 3rd, 3rd, you can make that label. No big deal. I'm going to keep it just on the scale of 1 and 2 and 3 like so. So then the vertex, I'm going to graph that first. Negative 1 third will be about right here. Negative 7 thirds, that's 2 and a 3rd, negative 2 and a 3rd. So it's going to be about right here. So the vertex, I'm going to graph it. I would put it right here and it's best to label this thing. So negative 1 third and then negative 7 thirds like so. Great. So we have the vertex there. We want the y-intercept. The y-intercept we can actually find from the original function. If you plug in x equals 0, you're going to get negative 2 as your y-intercept. So that's actually really close to negative 7 thirds. So draw it right here, right? So we're going to get 0, negative 2. And if we think of the axis of symmetry, the axis of symmetry always goes through the vertex of a parabola like so. What that does is for reflection purposes, you could reflect this on the other side to get another point, which would be about right here. Those are really still close together. So that's going to make, if I try to graph it from that, that would be very difficult. We do have to list the x-intercepts. I'll be explicit with the y-intercept. I put it on the graph, but I didn't list it. We need to do that. So the y-intercept is going to be 0, negative 2, like I mentioned, to find the x-intercepts. We have to set the equation equal to 0 and solve, for which actually the original format is probably a little bit easier to use to solve it. Notice my vertex turned out to be this fraction. Factoring probably wasn't going to be super helpful. We could, I mean, we could just solve from here. That's also an option. We could use the quadratic formula. There's so many options you could do here. You could just plug in the quadratic formula. I think for this one, I'm just going to take it, since I already put it into vertex form. Let's just go from there. Like, if you set this thing equal to 0, what's going to happen? So I'll switch my color here to just illustrate what's going on. If we're trying to solve f of x equals 0, that turns out to mean 3 times x plus a third squared is equal to 7 thirds. Divide both sides by 3. That's going to give us 7 over 9. We're going to take the square root of both sides. In which case, we end up with x plus a third is equal to, when we take the square, we're going to get plus or minus 7 over 3. And then we subtract the one third from both sides. Now we get our x-intercepts. Our x-intercepts, these are going to coincide with negative 1 plus or minus the square root of 7 over 3, for which you might need to approximate that one to help you out here. Again, using a calculator would be grand, but if you didn't have one, notice that the square root of 7 is almost the square root of 9, which means that thing's almost 3. And so again, from an estimate point of view, it's kind of like negative 1 plus 3 over 3. You get that as you get 2 thirds, and then the other one's kind of like negative 1 minus 3 over 3, which is going to give us negative 4 over 3. So roughly speaking, we're looking for negative 4 thirds and positive 2 thirds. That's not exactly the case. But again, if you have a calculator, you should just estimate it yourself. But if you didn't have one, you're not completely without any pursuit there. And so let's label these on the graph. And one was about 2 thirds right here, and we'll label it. This one right here would be negative 1 plus the square root of 7 over 3 comma 0. And then by symmetry, we get the other one, but that would be about right here, in which case this one was negative 1 minus the square root of 7 over 3 comma 0. And then from at some point, we're probably ready to start drawing our parabola, getting the correct concavity. It doesn't have to be perfect, but you should pass through your x-intercepts, your vertex, and your y-intercept. And then we get the picture like so. So we've graphed our quadratic function by first putting it in vertex form, identifying the intercepts in the vertex, and then graphing it.