Quadratic Functions - Online Tutor, Practice Problems & Exam Prep

Properties of Parabolas

Hey, everyone. We've 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 going to 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.

A quadratic function is a polynomial of degree 2 that has a standard form of f(x) = ax^2 + bx + c. Now all of these functions here 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 0 so long as a is not 0 and our largest exponent is still 2 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. We've worked with the square function before f(x) = x^2 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 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 going to look at the different elements of a parabola here and some are going to be 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. 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. 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 (0,0). Now for my other function here, it is not at the origin, it is actually at the point (-2,1), so my vertex is (-2,1). Now the other thing that I want to consider with my vertex is whether I'm dealing with a minimum or a maximum point. Looking at my square function here, this is the lowest point on my graph of that 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 (0,0), 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 (-2,1) and it is a maximum.

Let's look at something that we're more familiar with, our x intercept. The x intercept of a graph is anywhere that our graph crosses our x axis, not our y axis, our x axis. You'll only ever have 1 or 2 x intercepts, never more or less than that. Let's move on to our y intercept. 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 which again is at the origin, simply (0,0). And for my other graph, my y intercept is also at (-3,0).

We've looked at our intercepts and our vertex. Now we want to look at our axis of symmetry. 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 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=0, a vertical line through my vertex. And in our other function, if I draw a line that divides this perfectly in half, it again goes straight through that vertex point at x=-2.

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's always going to be negative infinity to infinity, which you might also recognize as being all real numbers. Now our range is going to depend on whether we have a minimum or a maximum point. So since we're dealing with a minimum with our square function, my range is from that minimum point at (0,0) all the way up to infinity, written as [0, \infty). For our other function, dealing with a maximum point at (-2,1), it's going to be from negative infinity up to my maximum point, written as (-\infty, 1].

Now we wanna look at one last thing on our graphs here. Something that you're often asked when dealing with parabolas is to tell them the interval for which your parabola is increasing and decreasing. For our second function that's facing downward and decide where it's increasing and decreasing. From negative infinity up to that vertex point at x=-2, it is increasing, written as (-\infty, -2) and from -2 all the way to infinity, we're going to be decreasing, written as (-2, \infty). 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 going to help us to easily graph these quadratic functions, which is what is coming up next.

Thanks for watching, I'll see you in the next video.

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

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

• A. $x=3$
• B. $x=0$
• C. $x=1$
• D. $x=4$

Vertex Form

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(x)=x2 and puts them all together in one equation that is then going to tell us exactly how to graph our quadratic function. So I'm going to 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(x)=a⋅(x-h)2+k and we have our a, h, and k that all represent transformations that are happening to our x2 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 has a negative, if my a is negative, it is instead going to open down. So I'm going to 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 1 so some number like 5 or 20 or anything greater than 1, that tells us that we are undergoing a vertical stretch. So if I take my x2 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 1, so something like 1/2 or 2/3 or any number less than 1, I am instead undergoing a vertical compression. So instead of stretching my x2 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-h, so we need to be careful with our signs there. X-h. Our 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 units. So if I have a quadratic function in vertex form, if I have my value h here, this is x-1, so one is my value for h. That tells me that I'm just going to shift 1 unit over and now my parabola is here. So it's shifted by some number h to the right. Now k is going to tell me my vertical shift. So instead of my horizontal, now it's dealing with a vertical shift by some number k units. Now here, I have x-1 2-4. So this minus 4 tells me my k value. So this tells me that I'm going to shift vertically by negative 4 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 4. 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 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, k. So here, we've already identified what our h and our k are, so we know that our vertex is simply going to be 1, -4. 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 2 is our axis of symmetry. Now our axis of symmetry is actually always going to be the line x=h, so just that point in our vertex. So here, my axis of symmetry is simply going to be x=1 and I'm done. I can move on to step 3. Now step 3 is to find the x-intercepts. Now this is where we're going to have to do a little bit more 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(x)=0. So if I take my function here and I set it equal to 0 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+somenumbers2=aconstant, 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-1 2-4. I'm going to go ahead and move that 4 over to the other side, canceling it out, leaving me with x-1 2=4. Now using the square root property, I'm going to go ahead and square root both sides, leaving me with x-1=±4, which is just 2. So from here, I can go ahead and move my one over by simply adding 1 to both sides, leaving me with just x=±2+1. So now I can just go ahead and split it into my 2 possible answers, positive 2 + 1 and negative 2 + 1. So going ahead and solving those, 2+1 I know is 3 and negative 2+1 is simply -1. So here I have my 2 x-intercepts, 3 and -1, and I've completed step number 3. My x-intercepts are 3 and -1. Moving on to step number 4, 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(0), which means I'm just plugging in 0 for x.\ So down here, I have f(0)=0-1 2-4. So I've really just taken my vertex form and plugged in a 0 for x. So let's go ahead and calculate that. Now 0 - 1 is simply -1, so this is just -1 2-4. Now -1 2 is 1 - 4. This will leave me with -3. So this is my y-intercept, -3. Okay. 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 1, -4, let's go ahead and plot that. So 1, -4, right here, and then my axis of symmetry, which we're going to represent with a dotted line, is simply the line x=1 that goes all the way through my vertex there. And then my x-intercepts, which are 3 and -1, plotting those on my x-axis, 3 and -1, and then finally my y-intercept at -3. Okay,\ now connecting this with a smooth curve, I know it needs to be in the shape of a parabola and here is my parabola. I want to 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 x2 function here. You'll notice that this is just the same thing but shifted one over and down 4 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.

Example 1

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)² plus 2. 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)² plus k. So identifying our vertex, we need to identify both h and k. Now here in my function, this is x+1, 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 1, and then k we have this positive 2. So the vertex point is negative 1, 2. 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 1, 2 and it is at a maximum.

Now let's move on to step number 2 and identify our axis 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 3, finding our x intercepts, steps 3 and 4 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 2 right at the bottom here. Let's go ahead and set up our equation for step 3 and set up f of x is equal to 0. So my function here, negative one half times (x + 1)² + 2 is equal to 0.

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 going to go ahead and move this 2 over to the other side. So subtracting 2 from both sides leaves me with negative one half times (x + 1)² is equal to negative 2. 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 2, canceling that one negative one half out and leaving me on this side with (x + 1)² is equal to 4. 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 + 1 is equal to plus or minus 2. Now from here, I want to go ahead and move that positive one over to the other side by simply subtracting it, so minus 1 on both sides, and I'm left with x is equal to negative one plus or minus 2, which I know I can split into my 2 possible answers. Negative one plus 2 gives me a positive one, and then negative one minus 2 is going to give me a negative 3. So here I have my 2 zeros or my 2 x intercepts, one and negative 3. So filling that in on my table up here, I have 1 and negative 3 as my x intercepts, and we can go ahead and move on to step 4 and find our y intercept by calculating f of 0, plugging 0 into my function for x.

So my function here, f of 0 negative one half times (0+1)² plus 2. Now I can simply compute this by simplifying negative one half. 0 + 1 is just 1, so this is 1² plus 2. Simplifying this further, ...

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.

Converting Standard Form to Vertex Form

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 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 going 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 \( f(x) = x^2 + 6x + 7 \), 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 \( x^2 \). I just have an invisible one in front of that \( x \) so \( a = 1 \), \( b = 6 \), and \( c = \text{this positive } 7 \). Let's go ahead and get started with step 1, which is going to be to factor out \( a \) of my first two terms. So since here \( a \) is just 1, really if I factor out 1, I'm just left with \( x^2 + 6x \) because that \( 6x \) isn't going to change if factoring out a 1. So this is going to leave me with the form \( a(x^2 + \frac{b}{a}x + c) \) because I've simply factored out an \( a \) there. So I've completed step number 1.

Let's move on to step 2, which is going to be to add and subtract \(( \frac{b}{2a})^2 \) inside of our parentheses. So this \( \frac{b}{2a} \) is going to show back up. Let's go ahead and calculate that here. So \( \frac{b}{2a} \), I know that \( b \) here is just 6, \( 2 \times a \) is just 1, and this is equal to \( \frac{6}{2} \) which is just 3. Now that's my \( \frac{b}{2a} \). I want to make sure and square it here before I add it. That's going to give me 9. So, here, I want to add 9 and subtract 9 inside of my parentheses. Now I can do that because I'm really just adding nothing if I add 9 and then take it right back away. So I've completed that first part of step 2. Let's look at the second part. Now for the second part of step 2, I want to move my subtraction times \( a \) to the outside of my parentheses. So I want to take this negative 9, and I want to move it. But since it's inside a parenthesis being multiplied by a factor of \( a \) I need to make sure I account for that. Now here \( a \) is just 1 so all I'm doing is taking minus 9 times 1 to the outside in order to cancel it on the inside there. Now that one isn't going to do anything this is really just minus 9 but remember that if your \( a \) isn't 1 it's going to get a little bit more complicated. So we've completed step 2.

Let's go ahead and rewrite this to make it a little bit cleaner. So I still just have my 1 on the outside, but that's not doing anything. So I just have \( x^2 + 6x + 9 \). I moved that negative 9 away. So this is \( +7-9 \) on the outside. Okay. So moving on to step 3. Now I want to factor this to \( (x + \text{this } \frac{b}{2a})^2 \) and then simplify it. So this \( \frac{b}{2a} \) we've seen it before and we know that it's just this 3 right here so this simply factors to \( (x+3)^2 \) because it is a perfect square trinomial as we know happens with completing the square. So we want to take this \( +7 -9 \) into account and I add my negative 2 to the outside of those parentheses leaving me with \( (x + 3)^2 - 2 \) and I've completed step 3.

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 going to leave this for practice that we'll do in just a little bit, but let's take one more look at our function here. \( (x + 3)^2 - 2 \). 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.

Example 2

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(x)=-2x2-4x+6. 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 -2, b is -4, and c is 6. Getting started with step 1, we want to factor a out of just the first two terms. We just identified a as -2. So, if I pull -2 out of my function out of those first two terms that leaves me with x2+2x. Now remember only your first two terms and don't forget to tack that constant on the end. Step 1 is done here.

Now I can go ahead and move on to step 2, which is to both add and subtract b over 2a squared inside of my parenthesis. So let's go ahead and compute b over 2a first. So b is -4, 2 times a, which is -2. This is -4 over -4, which is really just 1. Now if we take 1 and we square it, that just leaves us with 1. So we're going to add and subtract 1 here. Now -2x2+2x, and then a +1-1 inside of those parentheses, and then with my constant on the end. So that's the first part of step 2 is done.

We can go ahead and move on to the second part of step 2 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 2, I want to move it to the outside here. So this becomes negative one times negative 2, 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 -2 and then x2+2x+1. I moved my subtraction outside, so this just becomes +6, and then negative one times negative 2 is going to give me positive 2. So looking at all of this, I have finished step 2.

I can go ahead and move on to step 3, which is to factor 2x+b2a squared. So b over 2a, we already calculated, and we know that it's just 1. So this just becomes -2(x+1)2. That's what it factors down to. And then combining this 6 and 2 to simplify is going to give me a +8 on the end. And looking at this, I am done. I am in vertex form, and this is the function that I am now going to graph. So I can take my function here, and step 3 is done. I can move on to step 4 and go ahead and graph it right down here. So copying my function I have f(x)=-2(x+1)2+8. And we are ready to go. Let's get graphing.

Starting with step 1, we want to identify our vertex here, which is just h, k. Now looking at this, since I have x+1, 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 -1 and then k is just positive 8. So that's my vertex, -1, 8. Is this a minimum or a maximum point? Well, looking at my function here, I have this negative on the front, which tells me my parabola 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 -1. So this is simply the line x is equal to -1. Now moving on to steps 3 and 4, 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 3 and solve f(x) is equal to 0. So setting up our equation down here, my function is -2(x+1)2+8, and of course, equal to 0. So let's go ahead and come down here and solve this.

So I am going to go ahead and move my 8 over to the other side. Remember, we're going to be using the square root property with our vertex form here. So subtracting 8 from both sides, I have -2(x+1)2=-8. Then dividing both sides by -2 to isolate that squared expression, I have (x+1)2=4. Now I can go ahead and apply the square root property by square rooting both sides, leaving me with x+1=±4, which we know is 2. 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=-1±2.

Now here is, of course, where we want to split it into our 2 possible answers. So this is really -1 + 2 and -1 - 2. Now -1 + 2 is going to give me a positive 1, and the -1 - 2 is going to give me a -3. So these are my 2 x intercepts here, 1 and -3, which I can go ahead and fill in on my table up here, negative one and negative 3.

Moving on to our y intercept, we can go ahead and compute f(0) by plugging 0 into our function. So this is -2(0+1)2+8. Now simplifying this, I have -2(1)2+8. One squared is just 1, so this is -2 times one plus 8. -2 times one is, of course, just -2 plus 8. And then finally, our last step, -2 +8 gives me positive 6. So 6 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 symmetry right through that vertex. Then our x intercepts at 1 and at -3. So positive 1, negative 3, and then my y intercept at 6. So right up here. Okay. Now I can go ahead and connect all of this with a smooth curve. We know that this is a parabola facing 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.

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 8, including that 8 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 -1, it is increasing. So from negative infinity to -1, it is increasing. And then on the other side from -1 to positive infinity, it is falling back down, so it is decreasing. So decreasing from -1 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.

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

• A.
• B.
• C.
• D.