Quadratic Functions - Video Tutorials & Practice Problems

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1

concept

Properties of Parabolas

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7m

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Hey, everyone, we worked with some basic graphs of functions. And here we're going to focus on one specific type of function. The quadratic function. Now it might seem like a lot of information at first. And this graph behind me might look a little bit intimidating. But don't worry, I'm gonna walk you through absolutely everything that you need to know about the graphs of quadratic functions here. And you'll be an expert in no time. So let's go ahead and jump right in. So a quadratic function is a polynomial of degree two that has a standard form of F of X is equal to A X squared plus BX plus C. Now all of these functions are examples of quadratic functions and you can see that A B and C can be any real number, whether it be a fraction, a negative number or even zero. So long as A is not zero to E A is not equal to zero. And our largest exponent is still two making it a quadratic function. Now, here we're going to focus on the graphs of quadratic functions. So let's go ahead and take a look here. Now, we've worked with the square function before F of X is equal to X squared. And you may remember that this square function was a parabola. Now, this is actually going to be the same for all quadratic functions. They are all going to have this curved parabolic shape, whether it is a right side up upside down or located anywhere on our coordinate plane. Our quadratic functions will always be that same shape of a parabola. So we're gonna look at the different elements of a Parabola here and some are gonna be that were ones that we're familiar with like the X intercept or the Y intercept. But we're also going to work with some that are specific to parabolas like the vertex or the axis of symmetry. So let's go ahead and get started with our vertex. So the vertex of a Parabola is either going to be the lowest point or the highest point depending on whether our Parabola is opening upward or opening downward. So for our square function here, we see that our vertex is right at that origin point. And we're always going to write our vertex as an ordered pair. So my vertex is simply 00. Now, for my other function here, it is not at the origin, it is actually at the point negative to one. So my vertex is negative to one. Now, the other thing that I want to consider with my vertex is whether I'm dealing with a minimum or a maximum point. So again, looking at my square function here, this the lowest point on my graph of that uh quadratic function. So it is a minimum point. It's all the way at the bottom and nothing goes below it on that parabola. So not only is my vertex here, 00, it also represents a minimum. Now for my other function, my vertex is all the way at the top which tells me that I am not dealing with a minimum. I'm actually dealing with a maximum point here. So here my vertex is negative to one and it is a maximum. So let's go ahead and look at something that we're more familiar with our X intercept. Now, the X intercept of a graph is anywhere that our graph crosses our X axis, not our Y AX is our X axis. So we're going to look for the points that cross that X axis. And between my two functions here I have three different points that cross my X axis that represent my X intercepts. So let's take a look at our square function. Now, there's only one point that crosses the X axis and it is at my origin. So my X intercept is simply zero. Now looking at my other function, there are two points that cross the X axis. So I need to count for both of them negative three and negative one as my X intercepts. Now you'll only ever have one or two X intercepts never less or more than that. So let's move on to our Y intercept. Now, the Y intercept is where a graph crosses the Y axis this time the Y axis. So let's look at our square function first, we have this point, this represents our Y intercept. Now, our Y intercept here is again at the origin simply at zero. And for my other graph, my other function, my Y intercept is down here at negative three. So it's simply negative three. Now, we've looked at our intercepts and our vertex. Now we want to look at our axis of symmetry. And now our axis of symmetry is something that is going to be specific to parabolas. And it represents the line that divides our parabola perfectly in half. It is symmetric about that line. So our axis of symmetry is going to perfectly cut it in half. So looking at my square function, it's going to go straight through the middle and it's actually always going to go straight through our vertex. So here my axis of symmetry is simply the line X equals zero, a vertical line through my vertex. Now, looking at our other function here, if I draw a line that divides this perfectly in half, you're often going to see this actually written with a dotted line, but I've just highlighted them here so that it's easier to see. But here again, we're going straight through that vertex point at X equals a negative two. So it's always going to be a straight vertical line through our vertex. And that represents our axis of symmetry. If I were to fold my parabola in half at that line, it's going to perfectly match up because it is symmetric about that line. Now, something else to consider whenever we're looking at graphs of functions is always the domain and the range. So first looking at our domain, the domain of all quadratic functions is actually always going to be the same and it is always going to be negative infinity to infinity, which you might also recognize as being all real numbers. So for both my square function and for my other function here, they are both all real numbers negative infinity to infinity. Now our range is going to depend on whether we have a minimum or a maximum point. So looking first at our square function, we know that we're dealing with a minimum point here and that looking at that my Y values can go anywhere from that minimum and all the way up beyond that. So whenever I'm dealing with a minimum, my range is always going to go from Y minimum, whatever where that minimum point is all the way up to infinity. So in this case, since I start down at zero, my range is simply zero to infinity. Now when I'm dealing with a maximum point like I have here on my other graph, it's always going to go from my maximum point down to negative infinity. So an interval notation, I would write that as negative infinity up to my maximum point Y maximum. So in this case, it goes from negative infinity up to my Y point at Y equals one. And that's my domain and range. Now we want to look at one last thing on our graphs here. Something that you're going to often be asked when dealing with Parabolas is to tell them the interval for which your Parabola is increasing and decreasing. Now, this is just a fancy way of saying for what values of X is our graph going up and for what values of X is it going down. So let's just take a look at our second function here that's facing downward and decide where it's increasing and decreasing. Now, looking at the first side of this going down here all the way up to my vertex point, we see that our graph is going up, it is increasing. So the interval for which it is increasing is simply from negative infinity up to that vertex point at negative two. And our vertex is actually always going to divide the intervals for which we are increasing or decreasing. So the other side from negative to all the way to infinity, we're going to be decreasing because we see our graph is going down. So this is just simply the opposite negative to, to infinity. So that's all the stuff we need to know about Parabolas. One more thing that I want to mention here is that you might have noticed that my equation here doesn't quite look like it's in standard form. And that's actually because we're often going to see quadratic functions written in what's called a vertex form, which is what that is right there. And it's actually going to help us to graph with ease. It's going to help us to easily graph these quadratic functions, which is what is coming up next. So thanks for watching, I'll see you in the next video.

2

Problem

Problem

Identify the ordered pair of the vertex of the parabola. State whether it is a minimum or maximum.

A

$\left(-3,-2\right);$ maximum

B

$\left(-3,-2\right);$ minimum

C

$\left(-1,2\right);$ maximum

D

$\left(-1,2\right);$ minimum

3

Problem

Problem

Where is the axis of symmetry located on the given parabola?

A

$x=3$

B

$x=0$

C

$x=1$

D

$x=4$

4

concept

Vertex Form

Video duration:

8m

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Hey, everyone, by now, we've graphed some basic functions and we've even transformed those graphs by maybe shifting them some number of H units to the right or K units up. Or maybe even stretching it by a factor of A and the vertex form of a quadratic function takes all of those possible transformations that we could do to the square function F of X equals X squared and puts them all together in a one equation that is then going to tell us exactly how to graph our quadratic function. So I'm gonna walk you through step by step how to use this vertex form in order to easily graph our quadratic function using a bunch of stuff. We already know. Seriously, I'm not going to teach you anything new here. I'm just going to take a bunch of stuff we already know and bring it together. So let's go ahead and get started here and take a look at our vertex form. So here we have F of X is equal to A times X minus H squared plus K and we have our A H and K that all represent transformations that are happening to our X squared function. So let's first look at a, now we want to consider two things when looking at a, both the sign and the value of it. So first considering the sign, if A is positive, that tells us that our parabola is going to open upward. So like our square function, it's just going to open straight up. Now, if it, if it has a negative, if my A is negative, it is instead going to open down. So I'm gonna have a parabola that is opening downward like this. Now, the other thing we want to consider when looking at A is the actual value of it. So we're only considering the absolute value because we already took care of the sign. So if the absolute value of A is greater than one, so some number like five or two 20 or anything greater than one that tells us that we are undergoing a vertical stretch. So if I take my X squared function and I stretch it upward, it's going to look something like this much skinnier because I have stretched it. Now, if instead the absolute value of A is less than one. So something like one half or two thirds or any number less than one, I am instead undergoing a vertical compression. So instead of stretching my X squared function, I've instead compressed it and now it will look a little bit wider here. So that's a, let's go ahead and move on to H. So our value for H recognize that this is X minus H. So we need to be careful with our signs there. X minus hr H is going to tell us what we are horizontally shifting by. So H horizontal, it's a good way to remember. It is going to tell us our horizontal shift by some number H unit. So if I have an a quadratic function in vertex form, if I have my value H here, this is X minus one. So one is my value for H that tells me that I'm just going to shift one unit over. And now my parabola is here. So it shifted by some number H to the right now K is going to tell me my vertical shift. So instead of my horizontal, I'm not dealing with a vertical shift by some number K units. Now, here I have X minus one squared minus four. So this minus four tells me my K value. So this tells me that I'm going to shift vertically by negative for units paying attention to that sign there. So I'm just taking my parabola and I'm shifting it vertically by some number of units here by negative four. So we've looked at all of these different transformations. Let's go ahead and go through step by step how to graph a function in a vertex form. So the first thing that we want to consider is our actual vertex. It's in vertex form let's go ahead and find our vertex. So our vertex is actually always simply going to be the ordered pair H comma K. So here we've already identified what our H and RK are. So we know that our vertex is simply going to be one negative four. Now, the other thing we want to consider here is whether our vertex is going to be a minimum or a maximum point, which we can do by looking at our value for a here, I don't actually have a number for a which tells me that I'm dealing with an invisible positive one here because there's nothing else going on. So I know that A is positive, which tells me that my Parabola is going to open up. Now when my parabola opens up, that tells me that I am definitely dealing with a minimum here. So I know that that vertex will be a minimum. Now, the next thing we want to consider for step two is our axis of symmetry. Now, our axis of symmetry is actually always going to be the line X is equal to H. So just that point in our vertex. So here my axis of symmetry is simply going to be X equals one and I'm done, I can move on to step three. Now, step three is to find the X intercepts. Now this is where we're going to have to do a little bit or math. But don't worry, it's using something that we already know how to solve a quadratic equation. So here we're going to want to solve the equation F of X is equal to zero. So if I take my function here and I set it equal to zero, and I think about how to solve that, you might recall that we have a bunch of different methods. So if I consider everything that's happening here, and I'm looking for a clue of how I should actually solve this. I can see that this has the form X plus sum number squared equals a constant or rather I can make it in that form. So I know that I need to be using the square root property. So let's go ahead and solve this using the square root property. So I have X minus one squared minus four. I'm going to go ahead and move that four over to the other side, canceling it out, leaving me with X minus one squared is equal to four. Now using the square root property, I'm going to go ahead and square root both sides leaving me with X minus one is equal to plus or minus the square root of four, which is just two. So from here, I can go ahead and move my one over by simply adding one to both sides. Leaving me with just X is equal to plus or minus two plus one. So now I can just go ahead and split it into my two possible answers, positive two plus one and negative two plus one. So going ahead and solving those two plus one, I know is three and negative two plus one is simply negative one. So here I have my two X intercepts three and negative one. And I've completed step number three, my X intercepts are three and negative one. Moving on to step number four, which is to find our Y intercepts. Now we're going to have to do a little more math here. But all we're doing is computing F of zero, which means I'm just plugging in zero for X. So down here, I have F of zero is equal to zero minus one squared minus four. So I really just take in my vertex form and plugged in a zero for X. So let's go ahead and calculate that. Now zero minus one is simply negative one. So this is just negative one squared minus four. Now negative one squared is one minus four. This will leave me with negative three. So this is my Y intercept negative three. OK. So I found a bunch of points here. Now my last step to actually graph this is going to be to plot all of these points and then connect them with a smooth curve. We already know what the shape of our quadratic function should be a parabola. So we know what it should look like. Let's go ahead and plot all these points. So first starting with my vertex, one negative four. Let's go ahead and plot that. So one negative four right here. And then my axis of symmetry, which we're going to represent with a dotted line is simply the line X equals one that goes all the way through my vertex there. And then my X intercepts which are three and negative one, plotting those on my X axis three and negative one. And then finally my Y intercept at negative three. OK. Now connecting this with a smooth curve, I know it needs to be in the shape of a parabola. And here is my problem. I wanna put arrows on the end to represent that it doesn't just stop there. It keeps going and looking at this graph. So we're done, we've completely graphed it. I know that I have my X squared function here. You'll notice that this is just the same thing but shifted one over and down for as was shown in our vertex form. So even though we can always just sketch it from the beginning, we want to make sure and calculate these extra points just so that we can draw it accurately. So that's all you need to know in order to graph a function from vertex form. Let's get some practice.

5

example

Example 1

Video duration:

7m

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Hey, everyone, let's work through this example together. So here we want to graph the given quadratic function identifying all of the possible information about it. Now, here we have the function F of X is equal to negative one half times X plus one squared plus two. So let's get right into graphing. Now, looking at the very first thing I want to do, I want to identify my vertex which when written in vertex form is H comma K. Now just a reminder of vertex form, remember it's a times X minus H squared plus K. So identifying our vertex, we need to identify both H and K. Now here in my function, this is X plus one and I know in vertex form it's X minus H. So whenever we have a plus in that, we want to make sure and be careful and identify this as X minus negative one so that we know our H is not positive one, it is negative one here. So here my X value my vertex is negative one. And then OK, we have this positive two. So the vertex point is negative one comma two. Now, is this A minimum or a maximum point. Well, looking back at my function, it has this negative at the front which tells me that my Parabola is going to be opening downward, telling me that I have a maximum point as my vertex. So my vertex is negative 12 and it is at a maximum. Now let's move on to step number two and identify our act of symmetry, which is simply X is equal to H which we know that H is negative one. So this is simply the line X is equal to negative one. Now moving on to step three, finding our X intercepts steps three and four are going to be a little bit more involved because we need to calculate some stuff. So I'm going to be doing my work for these two right at the bottom here, let's go ahead and set up our equation for step three and set up F of X is equal to zero. So my function here negative one half times X plus one squared plus two is equal to zero. So looking down here at my function since I have something squared and a constant, I know that I'm going to go ahead and use the square root property. Whenever we have our function in vertex form, we can always solve using the square root property. So I'm gonna go ahead and move this two over to the other side. So subtracting two from both sides leaves me with negative one half times X plus one squared is equal to negative two. Now, from here, I want to go ahead and cancel out this negative one half which I can do by simply multiplying both sides by negative two, canceling that one, negative one half hour. And leaving me on this side with X plus one squared is equal to negative two times negative two, which will give me positive four. Now, from here, I can go ahead and apply the square root property and simply take the square root of both sides. Leaving me with X plus one is equal to plus or minus the square root of which we know is just two. Now, from here, I want to go ahead and move that ne that positive one over to the other side by simply subtracting it. So minus one on both sides and I'm left with X is equal to negative one plus or minus two, which I know I can split into my two possible answers. Negative one plus two. So moving this down here, negative one plus two and negative one minus two splitting into my two answers. Now negative one plus two gives me a positive one and then negative one minus two is going to give me a negative three. So here I have my two zeros or my two X intercepts one and negative three. So filling that in on my table up here, I have one and negative three is my X intercepts and we can go ahead and move on to step number four and find our Y intercept by calculating F of zero, plugging zero into my function for X. So my function here F of zero, negative one half times zero plus one squared plus two. Now I can simply compute this by simplifying negative one half zero plus one is just one. So this is one squared plus two. Simplifying this further negative one half, one squared is just one and then plus two, which simplifies further into negative one half plus two. Now I have a fraction and a whole number, we can go ahead and change this whole number into a fraction if that helps you out here. So this two is really just 4/2. So this is negative one half plus 4/2 which I found by just multiplying two times 2/2. Now, since these have a common denominator, I can easily add them together or subtract them since I have a negative here. So negative one half plus 4/2 gives me 3/2. And that is my intercept 3/2. I can go ahead and fill that in on my table as well 3/2. Now it's totally OK to get a fraction for anything, we're still going to be able to plot that. So now I have all of my information that I need in order to graph and I can go ahead and move on to step five and simply plot and then connect with a smooth curve. So let's go back up to our graph here. Now, looking at all of the information I have in my table, I'm gonna first plot my vertex at negative 12 and then my axis of symmetry at X equals negative one. I'm going to draw my dotted line right through that vertex point. Now for step three, I can go ahead and plot these X intercepts at one and negative three, positive one, negative three. And then finally, my Y intercept at 3/2, which is just 1.5. So it's in between one and two right here. OK. So I've plotted everything that I calculated. Now I can go ahead and connect all of this with a smooth curve. Now it's OK. If your curve isn't perfect, we know that this is a parabola facing downwards, your curve might not always look perfect. And that's totally OK. We're drawing this by hand. So that's all of our graph. Now we have that we can go ahead and find all of this remaining information. So first looking at our domain, we know that the domain of every single quadratic function is always going to be negative infinity to infinity or simply all real numbers. However, you want to write that or however your professor wants you to write that is totally OK. Here both are correct. Now, looking to our range, our range here since we have a maximum point is going to go from negative infinity until we reach our maximum point. Which here is Y equals two at that vertex. Remember that too is included in your range because I only have a point plotted there. So we want to make sure we use a square bracket on that range. Now we look for increasing and decreasing intervals. Remember that these are just for what X values is your graph going up for what X values is your graph going down. So looking at our graph here from negative infinity until I reach that vertex point at X equals negative one, my graph is going up. So it is increasing. So from negative infinity to negative one, it is increasing and then from negative one to positive infinity, my graph is going down or decreasing. So negative one to infinity. Now that negative one is never going to be included in these intervals because it is the point that it switches from increasing to decreasing. But it will always be your vertex point splitting that for our parabolas here. So that's absolutely everything that we need to do for this graph. And we found everything we need to. Thanks for watching and let's get some practice.

6

Problem

Problem

Graph the given quadratic function. Identify the vertex, axis of symmetry, intercepts, domain, range, and intervals for which the function is increasing or decreasing. $f\left(x\right)=-\left(x-5\right)^2+1$

A

B

C

D

7

concept

Converting Standard Form to Vertex Form

Video duration:

4m

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Hey, everyone, we just learned that whenever we have a quadratic function in vertex form, we can quickly find our vertex and our X and Y intercepts in order to easily graph it. But what if I'm given a quadratic function in standard form instead, how am I supposed to find my vertex? If I don't know what H and K are? Well, the good news is is that we can actually just take our standard form function and put it right back in vertex form by looking to a familiar friend completing the square. So here I'm to show you how you can take your standard form function, rewrite it and then add some number and subtract that same number in order to get back to vertex form by using something that you already know how to do, completing the square. So let's go ahead and get started here. Now, something important to consider is that we're now working with a function instead of an equation. So our steps are going to change slightly and you're going to need to pay attention and be careful with some of the changes that are happening. So let's go ahead and look at our example, here, we have F of X is equal to X squared plus six X plus seven. So we want to take this standard form and put it into vertex form. Let's go ahead and identify A B and C here before we get started. So here I just have X squared, I just have an invisible one in front of that X. So A is one, B is then six and C is this positive seven. Let's go ahead and get started with step one, which is going to be to factor out a of my first two terms. So since here A is just one, really, if I factor out one, I'm just left with X squared plus six X because that six X isn't going to change if factoring out a one. So this is going to leave me with the form A times X squared plus B over A times X plus C because I've simply factored out an A there. So I've completed step number one. Let's move on to step two, which is going to be to add and subtract B over two times A squared inside of our parentheses. So this B over two A is going to show back up. Let's go ahead and calculate that here. So B over two A I know that B here is just 62 times A is just one and this is equal to 6/2, which is just three. Now, that's my B over two A, I want to make sure in square here before I add it, that's going to give me nine. So here I want to add nine and subtract nine inside of my parentheses. Now, I can do that because I'm really just adding nothing if I add nine and then take it right back away. So I've completed that first part of step two. Let's look at the second part. Now for the second part of step two, I want to move my subtraction times A to the outside of my parentheses. So I want to take this negative nine and I want to move it. But since it's inside of parentheses being multiplied by a factor of A, I need to make sure I account for that. Now here A is just one. So all I'm doing is taking a minus nine times one to the outside in order to cancel it on the inside there. Now that one isn't gonna do anything. This is really just minus nine. But remember that if your A isn't one, it's gonna get a little bit more complicated. So we've completed step two. Let's go ahead and rewrite this to make it a little bit cleaner. So I still just have my one on the outside, but that's not doing anything. So I just have X squared plus six X plus nine. I moved that negative nine away. So this is plus seven minus nine on the outside. OK. So moving on to step three, now, I want to factor this to X plus B over two A, there's that number again squared and then simplify it. So this B over two A, we've seen it before and we know that it's just this three right here. So this simply factors two X plus three squared because it is a perfect square trinomial as we know happens with completing the square. So we wanna take this plus plus seven minus nine into account and I add my negative two to the outside of those parentheses, leaving me with X plus three squared minus two. And I've completed step three. Now, this form of an equation or form of a function might look familiar to you because it is now in vertex form and we can simply graph from vertex form. Now, I'm gonna leave this for practice that we'll do in just a little bit. But let's take one more look at our function here X plus three squared minus two. Now, I can easily extract H and K and graph it from there. That's all you need to know to take standard form back to vertex form. And now you can graph it just like, you know how, let's go ahead and get some practice.

8

example

Example 2

Video duration:

7m

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Hey, everyone, let's put everything that we just learned together in this example. So we want to graph the given quadratic function and then identify all of the possible information about our parabola. And the function I have here is F of X is equal to negative two X squared minus four X plus six. But seeing that this function is in standard form, I want to go ahead and convert it to vertex form by completing the square so that it's much easier for us to graph. So let's go ahead and do that here. The first thing I want to do is identify A B and C before I get started with my steps. So A here is negative two, B is negative four and C is six. So getting started with step one, we want to factor a out of just the first two terms. So we just identified A as negative two. So if I pull negative two out of my function out of those first two terms that leaves me with X squared and then plus two X. Now remember only your first two terms and don't forget to tack that constant on the end So step one is done here. Now I can go ahead and move on to step two, which is to both add and subtract B over two A squared inside of my parentheses. So let's go ahead and compute B over two A first. So B is negative four, two times A which is negative two, this is negative four over negative four, which is really just one. Now, if we take one and we square it, that just leaves us with one. So we're going to add and subtract one here. Now negative two X squared plus two X and then plus one minus one inside of those parentheses. And then with my constant on the end. So that's the first part of step two is done, we can go ahead and move on to the second part of step two which is going to be to move my subtraction times a outside of my parentheses. So here this negative one and then times this negative two. I want to move it to the outside here. So this becomes a negative one times negative two and that cancels it inside of my parentheses. So let's go ahead and rewrite this to get it a little more compact. So this negative two and then X squared plus two, X plus one, I moved my subtraction outside. So this just becomes plus six and then negative one times negative two is going to give me positive two. So looking at all of this, I have finished step two, I can go ahead and move on to step three, which is two factor two X plus B over to A squared. So B over two A we already calculated and we know that it's just one. So this just becomes negative two. And then inside of my parentheses, X plus one squared, that's what it factors down to. And then combining this six and two to simplify is going to give me a plus eight on the end. And looking at this, I'm done, I'm in vertex form. And this is the function that I am now going to. Yeah, so I can take my function here and step three is done. I can move on to step four and go ahead and graph it right down here. So copying my function I have F of X is equal to negative two times X plus one squared plus eight and we are ready to go. Let's get graphing. So starting with step one, we want to identify our vertex here, which is just H comma K. Now looking at this, since I have X plus one, I know that this plus is really minus a negative because remember we have X minus H so this is X minus negative one. So our value for H is negative one and then K is just positive eight. So that's my vertex negative one comma eight. Is this a minimum or a maximum point? While looking at my function here, I have this negative on the front which tells me my parable is opening downwards, which means my vertex is at the top, it's at a maximum. Then identifying our axis of symmetry here X is equal to H we just identified H as negative one. So this is simply the line X is equal to negative one. Now moving on to steps three and four, we know we're going to do some calculations which again I'm going to do right down here. Let's go ahead and start with step three and solve F of X is equal to zero. So setting up our equation down here, my function is negative two times X plus one squared plus eight and of course equal to zero. So let's go ahead and come down here and solve this. So I'm gonna go ahead and move my eight over to the other side. Remember we're going to be using the square root property with our vertex form here. So subtracting eight from both sides, I have negative two times X plus one squared is equal to negative eight. Then dividing both sides by negative two to isolate that squared expression I have X plus one squared is equal to negative eight divided by negative two, which is positive four. Now I can go ahead and apply the square root property by square rooting both sides. Leaving me with X plus one is equal to plus or minus the square root of four, which we know is two. Now, from here, I can go ahead and isolate X by moving my one over to the other side by subtracting, canceling out and leaving me with X is equal to negative one plus or minus two. Now here is of course where we want to split it into our two possible answers. So this is really negative one plus two and negative one minus two. Now, ne negative one plus two is going to give me a positive one and the negative one minus two is going to give me a negative three. So these are my two X intercepts here one and negative three which I can go ahead and fill in on my table up here negative one and negative three. OK. Moving on to our Y intercept, we can go ahead and compute F of zero by plugging zero into our function. So this is negative two times zero plus one squared plus eight. Now simplifying this, I have negative two times one squared plus 8, 1 squared is just one. So this is negative two times one plus eight, negative two times one is of course just negative two plus eight. And then finally our last step negative two plus eight gives me positive six. So six is going to be our Y intercept here filling that in on my table. Now we are done, we can go ahead and plot and connect everything with a smooth curve. So coming right back up to my graph here, let's start by graphing our vertex point negative 18. So negative one positive eight. And then of course, putting our axis of symmetry right through that vertex, then our X intercepts at one and negative three. So positive one negative three and then my Y intercept at six. So right up here, OK. Now I can go ahead and connect all of this with a smooth curve. We know that this is a parabola facing it downwards because we identified our maximum point. So here is my parabola and now I can go ahead and look at all of this remaining information and look at my graph to fill that in. So of course, our domain still the same still negative infinity to infinity, nothing changed there. And then our range, since we're dealing with a maximum point, we know goes from negative infinity up into that Y max which in this case is eight, including that eight in a square bracket. Of course, then finally, we have increasing and decreasing. Increasing is for what X values is that going up? Which from negative infinity until I reach that vertex point at X equals negative one, it is increasing. So from negative infinity to negative one, it is increasing. And then on the other side of my parabola from negative one to positive infinity, it is falling back down. So it is decreasing. So decreasing from negative one to positive infinity, those increasing and decreasing intervals always separated by that vertex. So that's all for this one, I'll see you in the next one.

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Problem

Problem

Graph the given quadratic function. Identify the vertex, axis of symmetry, intercepts, domain, range, and intervals for which the function is increasing or decreasing. $f\left(x\right)=3x^2+12x$