Quadratic Functions - Online Tutor, Practice Problems & Exam Prep

Quadratic functions, expressed in standard form as f(x) = a(x-h)² + k, exhibit a parabolic shape. Key features include the vertex, axis of symmetry, and intercepts. The vertex indicates minimum or maximum points, while the axis of symmetry is x = h. Understanding transformations through parameters a, h, and k aids in graphing.

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.

So a quadratic function is a polynomial of degree 2 that has a standard form of \( f(x) = ax^2 + bx + c \). 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 \( a \neq 0 \) 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. Now we've worked with the square function before \( f(x) = x^2 \) and you may remember that this square function was a parabola. 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 the 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.

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 (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). 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 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.

Now let's go ahead and look at my x intercept. The x intercept of a graph is anywhere that our graph crosses our 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 \(0\). Now looking at my other function, there are two points that cross the x axis, so I need to account for both of them, -3 and -1, as my x intercepts.

Now let's move on to our y intercept. The y intercept is where a graph crosses 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 \(0\). And for my other graph, my other function, my y intercept is down here at -3, so it's simply \(y = -3\).

Now we've looked at our intercepts and our vertex. Now we want to look at our axis of symmetry. Our axis of symmetry 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 \(x = 0\), 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 \(x = -2\). 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.

Somet

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. Let's first look at a. Now we want to consider 2 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. Now if a is negative, it is instead going to open downward.

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, that tells us that we are undergoing a vertical stretch. Now if instead the absolute value of a is less than 1, we are undergoing a vertical compression. 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. Our h 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 minus 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.

Now k is going to tell me my vertical shift. Here, I have (x−1)2−4. This minus 4 tells me my k value. So this tells me that I'm going to shift vertically by negative 4 units. 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. 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 (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 upward. Now when my parabola opens up, that tells me that I am definitely dealing with a minimum here.

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 is equal to h, so just that point in our vertex. So here, my axis of symmetry is simply going to be x equals 1. 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) is equal to 0. We know that we need to be using the square root property. So let's go ahead and solve this using the square root property. (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 minus 1 is equal to plus or minus the square root of 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 is equal to 3 and -1. So here I have my 2 x intercepts, 3 and -1.

Moving on to step 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. Now -1 squared is 1 minus 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. First, starting with my vertex (1, -4), let's go ahead and plot that. 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, 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. 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 \( f(x) = -\frac{1}{2}(x + 1)^2 + 2 \). 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 \( h, k \). Now just a reminder of vertex form, remember it's \( a(x - h)^2 + k \). 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 \( x-h \). So whenever we have a plus in that, we want to make sure and be careful and identify this as \( x - (-1) \) 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 \( (-1, 2) \). 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 \( (-1, 2) \) and it is at a maximum. Now let's move on to step number 2 and identify our axis of symmetry, which is simply \( x = h \), which we know that \( h \) is negative one. So this is simply the line \( x = -1 \). 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 \( f(x) = 0 \). So my function here, \( -\frac{1}{2}(x + 1)^2 + 2 = 0 \). 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 \( -\frac{1}{2}(x + 1)^2 = -2 \). Now from here, I want to go ahead and cancel out this \( -\frac{1}{2} \), which I can do by simply multiplying both sides by \( -2 \), canceling that \( -\frac{1}{2} \) out and leaving me on this side with \( (x + 1)^2 = 4 \). 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 \( x + 1 = \pm 2 \). Now from here, I want to go ahead and move that positive one over to the other side by simply subtracting it, so \( -1 \) on both sides, and I'm left with \( x = -1 \pm 2 \), which I know I can split into my 2 possible answers. \( -1 + 2 \) and \( -1 - 2 \), splitting into my two answers. Now \( -1 + 2 \) gives me a positive one, and then \( -1 - 2 \) 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 number 4 and find our \( y \)-intercept by calculating \( f(0) \), plugging 0 into my function for \( x \). So my function here, \( f(0) = -\frac{1}{2}(0+1)^2 + 2 \). Now I can simply compute this by simplifying \( -\frac{1}{2}(1)^2 + 2 \). Simplifying this further, \( -\frac{1}{2} + 2 \). 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 2 is really just \( \frac{4}{2} \). So this is \( -\frac{1}{2} + \frac{4}{2} \), which I found by just multiplying 2 times \( \frac{2}{2} \). Now, since these have a common denominator, I can easily add them together or subtract them since I have a negative here. So \( -\frac{1}{2} + \frac{4}{2} \) gives me \( \frac{3}{2} \), and that is my \( y \)-intercept, \( \frac{3}{2} \). I can go ahead and fill that in on my table as well. \( \frac{3}{2} \). Now it's totally okay 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 and I can go ahead and move on to step 5 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 \( (-1, 2) \), and then my axis of symmetry at \( x = -1 \). I'm going to draw my dotted line right through that vertex point. Now for step 3, I can go ahead and plot these \( x \)-intercepts at 1 and negative 3, positive 1, negative 3, and then finally my \( y \)-intercept at \( \frac{3}{2} \), which is just 1.5, so it's in between 1 and 2. Right here. Okay. 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 okay 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 okay. We're drawing this by hand. So that's all of our graph. Now we have that, and 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 okay 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 \( y = 2 \) at that vertex. Remember that 2 is included in your range because I only have a point plotted there, so we wanna make sure we use a square bracket on that range. Now we look for our 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 \( x = -1 \), my graph is going up, so it is increasing. So from negative infinity to \( -1 \), it is increasing. And then from \( -1 \) to positive infinity, my graph is going down or decreasing. So \( -1 \) to infinity. Now that \( -1 \) 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.

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 into 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)=x2+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 squared. I just have an invisible one in front of that x, so a is 1, b is then 6, and c is 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 x2+6x because that 6x isn't going to change if factoring out a 1. This is going to leave me with the form a times xx2 plus b over a times x plus 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 b over 2 times a squared inside of our parentheses. So this b over 2 a is gonna show back up. Let's go ahead and calculate that here. So b over 2 a, I know that b here is just 6, 2 times a is just 1, and this is equal to 6 over 2 which is just 3. Now that's my b over 2 a. I wanna make sure and square it here before I add it. That's gonna give me 9. So, here, I wanna 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 wanna 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. I still just have my 1 on the outside, but that's not doing anything. So I just have xx2+6x+9. I moved that negative 9 away. So this is plus 7 minus 9 on the outside. Okay. Moving on to step 3. Now I want to factor this to x+2a, there's that number again, squared, and then simplify to x+3 squared because it is a perfect square trinomial as we know happens with completing the square. So we wanna take this plus 7 minus 9 into account and I add my negative 2 to the outside of those parentheses, leaving me with x+3 squared minus 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 squared minus 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. Starting 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 and then a x. Remember, only your first two terms and do not forget to tack that constant on the end. Step 1 is done here.

Now I can move on to step 2, which is to both add and subtract b2×a squared inside my parentheses. So, let's compute b2×a first. b is -4. 2×a, which is -2. This is -4 over -4, which is really just 1. Now, if we square 1, that just leaves us with 1. So we're going to add and subtract 1 here. -2x2+2x, and then +1-1 inside those parentheses, with my constant on the end. That's the first part of step 2 done. We can move on to the second part of step 2, which is going to be to move my subtraction×a outside of my parentheses. So, here, this -1 and then times this -2, I want to move it to the outside here. So this becomes -1×-2, and that cancels it inside of my parentheses. 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 -1×-2 is going to give me positive 2. So looking at all of this, I have finished step 2. I can move on to step 3, which is to factor 2x+b2×a squared. We already calculated b2×a, and we know that it's just 1. So this just becomes -2. And then inside my parentheses, x+1. 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'm done. I'm 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-h. So this is x--1. 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? 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=-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)=0.

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 solve this. I'm going to move my 8 over to the other side. Remember, we're going to be using the square root property with our vertex form here. 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=-8÷-2, which is positive 4. Now I can go ahead and apply the square root property by squaring 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 1 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, -1 and -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 -2×12+8. One squared is just 1, so this is -2×1+8. Negative 2 times 1 is, of course, just

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.

Here’s what students ask on this topic:

What is the standard form of a quadratic function?

The standard form of a quadratic function is given by:

$f\left(x\right)=a{x}^2+bx+c$

where $a$, $b$, and $c$ are real numbers, and $a$ ≠ 0. This form is useful for identifying the coefficients that affect the shape and position of the parabola on the coordinate plane.

How do you find the vertex of a quadratic function in standard form?

To find the vertex of a quadratic function in standard form $f\left(x\right)=a{x}^2+bx+c$, use the formula:

$x=-b/\left(2a\right)$

This gives the x-coordinate of the vertex. To find the y-coordinate, substitute this x value back into the original function:

$y=f(-b/(2a\left)\right)$

The vertex is then $(-b/(2a),f(-b/\left(2a\right)\left)\right)$.

What is the vertex form of a quadratic function?

The vertex form of a quadratic function is:

$f\left(x\right)=a(x-h{)}^2+k$

where $a$ determines the direction and width of the parabola, and $h$ and $k$ are the coordinates of the vertex. This form is particularly useful for graphing because it directly shows the vertex and the transformations applied to the basic parabola $y=x^2$.

How do you convert a quadratic function from standard form to vertex form?

To convert a quadratic function from standard form $f\left(x\right)=a{x}^2+bx+c$ to vertex form $f\left(x\right)=a(x-h{)}^2+k$, follow these steps:

1. Factor out $a$ from the first two terms.

2. Complete the square inside the parentheses.

3. Add and subtract the same value inside the parentheses, then move the subtracted value outside.

4. Simplify to get the vertex form.

This process involves algebraic manipulation to rewrite the quadratic in a form that reveals the vertex directly.

What is the axis of symmetry in a quadratic function?

The axis of symmetry of a quadratic function is a vertical line that passes through the vertex of the parabola. For a quadratic function in vertex form $f\left(x\right)=a(x-h{)}^2+k$, the axis of symmetry is given by the equation:

$x=h$

This line divides the parabola into two mirror-image halves, ensuring that the function is symmetric about this line.