Algebra 1 Unit 6A Practice Test Form 2

Now I know from my experiences of graphing parabolas that as long as this is a one right here, then it's going to graph just like I'm doing right here. And I went over one and up one over another one and up three, or you could say I'm going over two and going up for that is how I get my nice parabola. That gives me 5 points. And I can also do that by making an in out table. And I'm going to put the vertex right in the middle. So that means I'm going to need zero and negative one. And two, and three. Now, if I start plugging those in up here, I'm going to put a two in here first. That's going to be two minus one, which is one. One squared is one plus two is three. Which means this is also three. If I put a three in here I get three minus one, which is two. Two squared is going to give me four, four plus two would be 6. And so those are both 6 and if you'll notice that was the point. 

So I went through the domain for all these parabolas is all real numbers. The range, however, I'm not going to use every Y in the book. I'm only using them from here up. So that's where Y is two, so it's going to be Y is greater than or equal to two. And you'll notice that it keys in on that Y coordinate of the vertex. So that's important. Well, the domain, in this case, I know is all real numbers. When I need to graph the Snell, whenever it's just a naked X squared like that, and it doesn't have an X term, I know that the X coordinate of the vertex is going to be zero. All this is is if I was talking about X squared, it's been stretched and dropped down four points. It's been stretched vertically and dropped down four points. That means it hasn't been shifted left or right, so that means it's going to have a vertex of zero for the X coordinate. Now, if I put a zero in there, I get two times zero squared minus four, that's going to get me a negative four. So yeah, that makes sense. It's been shifted down for units, so it's going to be right there. 

Now normally I'd put a point there and there, but because of the two, it's going to get stretched. So it's going to be somewhere else. So I should probably make a table. And in the middle of it, I'm putting zero negative four, which means before that would be negative one and negative two. And after that would be one and two. So let's try a one, two times one squared minus four one squared is one, so it's two minus four, which is negative two. And that makes sense because it's going to be stretched so it's going to be pulled up twice as fast as it normally would be. Let's try a two in there two times two squared minus four. Well, that's two times four minus four, so that's 8 minus four, which is four. So I'm going to go two units over and four units up and it is really stretched, isn't it? It's going to look something like this. All right, I already know what the domain is. The range on the other hand starts off here and goes up, so it's Y is greater than or equal to negative four. Okay. No, I can look at this one. And it's talking about some of these are features right of this thing. 

So I want to think about where those points are versus the vertex the vertex in this case is the lowest point. So that's a minimum. And it looks like negative two negative 9, which answers this question here. It is a minimum in its negative two negative 9. The X intercepts appear to be right here, which is one zero, and right here, which is a negative 5, zero. The Y intercept right there is at zero negative 5. The domain is just like those other ones. It's all real numbers. Because there's no X I can't put into this. All X's are going to show up on those X axis. There will always be a point. I can put on this graph. The range, on the other hand, has a limit. It can't be below negative 9. So why it's got to be greater than or equal to negative 9. Intervals on which it is increasing. Well, right here, it's decreasing. But from here on, it's increasing. So in terms of the X axis, this is the part I'm talking about. But it's increasing when X is greater than or equal to negative two. What you'll notice that has something to do with the vertex, doesn't it? Just like the range has something to do with the Y coordinate, the intervals on increasing and decreasing depend on the X coordinate of the vertex. So I know this is also going to involve negative two. 

The decreasing portion is this whole part here, and remember it goes on forever that way. So that means it's this part of the X axis. So starting at negative two and going to the left, so it's X is less than or equal to negative two. A function that represents two X squared, F of X equals two X squared after a translation left of 5 units. Well, in the standard not standard in the vertex form, we get something that looks like this. We know that the transformations that can take place a either is going to flip it if it's negative or if it's positive. It goes opens upward. If it's a number less than one, it's going to cause it to compact together vertically. Squeeze it together. If it's a number greater than one, it's going to cause it to vertically stretch and that's all for a, H causes a shift left or right. If it says minus some number, it's going to shift to the right. If it's plus some number, it's going to shift to the left because that negative tells me it works backwards for the X's. And then on the K, which has to do with the Y's, it's going to shift it up or down if it's a positive. It shifts up. If it's a negative, it shifts down. 

So in this case, all we've got going on right now, it's still going to vertex zero, zero. The two is going to cause it to stretch vertically. But after a translation of 5 units to the left, that's going to be the H so it's going to look like this F of X equals two times X let's see, left, so that's a plus 5 squared. And that causes that shift. Compared contrast these functions. Well, F of X because it's going to negative a, it's going to kind of open downward F of X opens downward. So does G of X because it's got that negative there. But it's also stretched vertically. This plus one causes F of X is shifted. To the right. Now, to the left, because that would be minus a negative to the left, one unit. G of X isn't shifted left to right, is shifted. That's the K right there, because it's not in the parentheses. Is shifted. Up to units. Okay, and that covers the whole thing. State the vertex. Well, that's easy. Vertex remember it's the opposite of this one. So the vertex is at. Negative two 7. And now the axis of symmetry remember the AOS is X equals in this case negative two because it's always going to be equal to whatever that X coordinate is the axis of symmetry always goes right through the vertex. 

So that's why that axis of symmetry is X equals negative two. Now we're going to convert this to standard form which won't have any parentheses. So the first thing I have to do is square that X plus two. So that's going to be, according to what we've learned from our boxes X squared plus four X plus four. And that's what you get when you take X plus two times X plus two, you get X squared to X to X and four. Those two add up to four X and then don't forget I could add 7 at the end. Now I need to distribute that negative three, so it's going to be negative three X squared -12 X. Don't forget to distribute it to the whole thing. -12, and then I'm adding that 7 at the end. So finally, the standard form is negative three X squared -12 X -5. That's the answer I was looking for. Okay, now this question is very similar to the one that was on the form one. First job I want to do is identify the vertex, which is going to be 5 32. Remember it's the opposite of this. And then just keep that the same. What does it mean? It means that in 5 seconds of the ball. 

Reaches its maximum. High, which is 32 feet. And since that's our vertex one as we'll graph at 5, 32 is about right there. And then if that's 5, 32, that's the vertex. It's in the middle. I'm going to need some points before it. And I need some points after. Okay, so for points before it and after. So let's try using a 6 in here. I'm going to do this on the back of another piece of paper so that I can do the calculations easily. So if I put a 6 in here, I'm going to get negative two times one squared, which is negative two, plus 32 is 30. So it's 6. I get 30. Which means that four, it's going to be 30. And I could get the three and the two, but what I really want to do is the one. So negative two times one -5 squared. Plus 32. One -5 squared is going to be negative four squared. Negative four squared remember is going to be a positive 16 because a negative times a negative is a positive. That's going to give me negative 32. This is going to be zero at one. It's zero, one at zero, so at 9, it is also zero. 

Now, there are some other values I could get here. I don't want to mess with it, but it makes sense to do them if you're having trouble figuring out where things are. So these are both going to be at 30. I learned that. And then it hits zero at one and 9. So it's going to look something like this. Okay, and that's half of X now the next thing it says is suppose you throw the ball three seconds later. Well, if it's three seconds later, this whole thing is going to shift three points over. One, two, three, one, two, three, one, two, three, one, two, three. And that's G of X and it's vertex right there is that 8 32, which allows me now to write the equation, it looks just like this one. Except instead of a 5, it's going to have an 8. And it kind of makes sense if it's two seconds, three seconds later. It's going to be 8. And there's my equation. Finally, this is explain the key features of a graphic of a quadratic function. And what they represent in real in a real world situation. 

Now, I like that Kathy example that was given to us on that real world's problem. And it gives us four points that are critical points. One was right there. So when we look at the cavities situation. A diver, the Y intercept. Tells us. When and where she's started. At zero seconds. She was on the top of a cliff. The X intercepts. Tell us when her output, in this case it was the height. Zero. So she entered the water. At one second. And exited. At 6 seconds. The vertex. Tells us the maximum. Or minimum point output. And Cathy's case. She reached. The lowest point in the water. At the vertex. It also tells us the time that she reached that point because that would be the X coordinate of that point. So if we said it was three negative 20, then at three seconds. She was 20 feet underwater. Like that.