Note the other Quadratics section that follow: Solving Quadratics by Factoring and Completing the Square, Quadratic Inequalities, and Quadratic Applications.
Here is a review in the next section that might help: Quadratic Review.
Quadratics and the Parabola
One special type of polynomial equation that you’ll work with a lot is a Quadratic Equation, (also known as a Quadratic Function, if it meets the requirements of being a function).
A quadratic is a polynomial that has an $ \boldsymbol{{x}^{2}}$ in it; it’s as simple as that. The degree of quadratic polynomials is two, since the highest power (exponent) of $ x$ is two. The quadratic curve is called a parabola. Technically, the parabola is the actual picture of the graph (shaped like a “$ \bigcup $”), and the quadratic is the equation that represent the points on the parabola. But a lot of times we hear the words “quadratic” and “parabola” used interchangeably.
This is interesting: if you were to create a chart with the $ x$’s and $ y$’s with a quadratic equation, if the $ x$’s are all spaced evenly apart, the second difference (the differences of the differences) of the $ y$’s are equal. (A linear equation has the first differences equal, as we saw earlier). For example, for $ y={{x}^{2}}$,
It seems like every math book talks about quadratics with either a thrown ball or rocket ship example. I’d like to change things up a bit and talk about a bride throwing her bouquet behind her – going up in the air and coming back down to be caught. We’ll look more at this specific example below.
For the purposes of this section, we will only deal with “vertical parabolas” in quadratics, that either go up or back down (like our bouquet), or they go down and then back up. (Actually, they can go sideways, too, as we’ll see in the Conics section, but these are not functions). Thus, these quadratics either have a maximum point, or a minimum point; whatever this point is, it’s called the vertex.
The basic (parent) function for a quadratic is $ y={{x}^{2}}$, where the vertex is $ (0,0)$ and the parabola faces up. There are many different forms of the transformed function (which we’ll talk about in the Parent Functions and Transformation section), as shown below. (The “sideways” parabolas are in the form $ x={{y}^{2}}$; these are not functions, since they don’t meet the vertical line test).
Below are the three basic equations for a parabola; Standard Form, Vertex Form, and Factored (Intercept) Form. They all tell us different things about the parabola. Do you see why they are called what they are called?
Here are some examples of parabolas in standard form, with their characteristics:
Graphing Quadratics (Parabolas)
Parent Function
Before we talk specifically about the Vertex and Factored forms of quadratics, let’s graph the simplest form of a parabola, $ y={{x}^{2}}$, using a t-chart.
Note that parabolas have symmetry; it is a mirror image of itself across the vertical line that contains the vertex; this line is called the line of symmetry (LOS), or axis of symmetry (AOS). In this case, our LOS is $ x=0$. Do you see how the LOS is always “$ x=x \text{ value of the vertex point}$” for these “up and down” parabolas?
You’ll probably be asked to get the domain (all the possible $ x$-values) and range (all the possible $ y$-values) when you graph parabolas. Remember to use your pencil to check the domain and range like we did here in the Algebraic Functions section.
For the parent function $ y={{x}^{2}}$, the domain is $ \left( {-\infty ,\infty } \right)$, or all real numbers, since these types of parabolas go up and out to the sides forever (this is true for every vertical parabola). However, the range is just $ \left[ {0,\infty } \right)$, since the parabola has a vertex at $ (0,0$) and the $ y$-values never get below 0.
Graphing with Vertex Form
It’s very important to find out how to get the vertex of a parabola, since that you way can get the line of symmetry (LOS) and graph it more easily. As we saw earlier, one of the forms of a quadratic equation is what we call vertex form. If we get the equation in that form, we can get the vertex easily:
With this equation, the vertex is the point $ (h,k)$. In the example above, the vertex is $ (-2,1)$. The LOS is $ x=-2$, and the vertex is a minimum, since there is no negative sign before the 3. Note that we have to be a little careful with the signs in this equation and remember that this equation is the same as $ y=3{{(x-(-2))}^{2}}+(-1)$, and that’s why the vertex is $ (-2,-1)$.
Let’s graph the equation $ y=3{{(x+2)}^{2}}-1$. Since the $ y$-intercept is when $ x=0$, we can easily solve for it: $ y$-intercept is $ y=3{{(0+2)}^{2}}-1=3{{(2)}^{2}}-1=12-1=11$; the point is $ (0,11)$.
We can graph the vertex and the $ y$-intercept, and then do a quick t-chart to find some of the other values. The graph is shifted down 1, over to the left 2, and a little skinnier than the graph of $ y={{x}^{2}}$. We’ll talk about these transformations of graphs later in the Parent Graphs and Transformations section.
The domain of this graph is still all real numbers, yet the range is $ \left[ {-1,\infty } \right)$; this is because the $ y$-part of the vertex is –1, so that is the lowest point of the range. Here is the graph:
One more thing to mention: we call points of a parabola that are directly opposite each other (over the LOS) a point of reflection, or reflection point. For example, the reflection point of $ (0,11)$ is $ (-4,11)$. See how both points are equidistant from $ x=-2$?
Also note that sometimes the equation is almost in vertex form; it might look something like $ \displaystyle y=-2{{\left( \frac{1}{2}x+2 \right)}^{2}}-3$, and we can do some algebra to change it into
$\displaystyle \begin{align}{l}y&=-2{{\left( {\frac{1}{2}\left( {x+4} \right)} \right)}^{2}}-3=-2{{\left( {\frac{1}{2}} \right)}^{2}}{{\left( {x+4} \right)}^{2}}-3\\&=-2\left( {\frac{1}{4}} \right){{\left( {x+4} \right)}^{2}}-3=-\frac{1}{2}{{\left( {x+4} \right)}^{2}}-3\end{align}$, so the vertex is $ (-4,3)$.
Graphing with Standard Form
When the equation is given in standard form, it’s useful to know the formulas below to get the vertex and $ y$-intercept. (We can also Complete the Square to go from Standard Form to Vertex Form, which we learn how to do here in the Solving Quadratics by Factoring and Completing the Square section.)
Once you graph the vertex and know which direction the parabola points, you can either graph the $ y$-intercept, or other points close to the vertex to graph the actual graph.
Graphing with Factored (Intercept) Form
With factored, or intercept form, we automatically have the $ x$-intercept(s), so we can graph those right away. Then, when we get the $ x$-intercepts, we can take the average (the point right in the middle) to get the Line of Symmetry (LOS). Then, to get the vertex, we can plug in the line of symmetry $ x$-point in the original equation to get the $ y$.
Standard Form to Vertex Form
Let’s look at some examples where we get the vertex from a standard form equation. Remember that when the coefficient of $ {{x}^{2}}$ is positive, the parabola faces up, and when the coefficient is negative, it faces down.
Let’s graph the last equation above $ y=-{{x}^{2}}+4x+3$, where the $ y$-intercept is $ (0,3)$. and the vertex is $ (2,7)$. The t-chart includes other points.
Example Real-Life Quadratic Application
Now let’s show how quadratics can be really useful in real life and model the flight of a bride’s bouquet. 
Let’s make the $ x$-axis be the time (in seconds) and the $ y$-axis the height of the bouquet at that time (in feet).
Note: The $ x$-axis is time, not a distance; sometimes parabolas represent the distance on the $ x$-axis and the height on the $ y$-axis, and the shapes are similar. Height versus distance would be the actual path or trajectory of the bouquet, whereas this graph is the height versus the time.)
Our axes will look like this: 
Let’s say, hypothetically speaking, that the trajectory of the bouquet (the height it takes in the air, per a certain amount of time) can be modeled by the equation below. We can also call this the “projectile” of the bouquet (usually in math books, they talk about trajectories of rockets or balls):
$ f(t)=-16{{t}^{2}}+20t+5$ Remember function notation? If not go back and review here in the Algebraic Functions section.
Why the parabolic shape? Think about it; if you were to throw something up in the air, it doesn’t keep going; it makes a curve and comes backs down. This is because of gravity.
Just for some history and a tiny bit of physics: The Italian scientist Galileo discovered the parabolic trajectory of projectiles in the early 17th century. Neglecting air friction, the $ -16{{t}^{2}}$ represents the gravity of of an object in feet per second per second (actually the $ 16$ is half the force of gravity). The constant ($ 5$ feet, in our example) represents the height $ h$ at which the object is thrown at the beginning, and the $ 20$ in the $ 20t$ represents the initial velocity $ {{v}_{0}}$ vertically upward, feet per second, or how fast the object is thrown up in the air. Thus, we have $ f(t)=-16{{t}^{2}}+{{v}_{0}}t+h$ for these applications.
Let’s figure out how high the bouquet goes, when it hits this highest point, and when it hits the ground, assuming it is not caught.
To get the highest point of the bouquet, we need to get the vertex, which will be at a maximum, since we have a negative coefficient before the $ {{x}^{2}}$. Since our quadratic equation is in standard form, let’s use our cool $ \displaystyle \left( {-\frac{b}{{2a}},\,\,f\left( {-\frac{b}{{2a}}} \right)} \right)$ trick to obtain the vertex (highest point), and $ y$-intercept to graph our equation:
The highest point is at (.625 seconds, 11.25 feet). Our $ y$-intercept is $ (0,5)$.
Here’s what the graph looks like. Note that even though this may look like the path of the bouquet (distance of the bouquet from where it was thrown), we are actually graphing the height of the bouquet per time.
Note that we’ll see this type of application, as well as more applications in the Quadratic Applications section.
Solving for Roots with Quadratics
Many times we want to know where the quadratic hits the $ \boldsymbol{x}$–axis (for example, to find out where the bouquet in our example hits the ground); these are called “$ \boldsymbol {x}$-intercepts“, “zeroes“, or “roots” of the quadratic equation. This is how we “solve” quadratic equations; we set them to 0 (like we do with linear equations!). Quadratics can have one or two roots, and when they never hit the $ \boldsymbol{x}$-axis the roots are “imaginary”; we’ll address them later in the Imaginary (Non-Real) and Complex Numbers section.
Again, as we saw above, when the quadratic is in factored form, we can get the $ x$-intercepts by setting each factor to 0 and solving. We will learn about other ways to solve quadratic equations in the Solving Quadratics by Factoring and Completing the Square section.
Quadratic Formula (Quadratic Equation) for Solving Roots
When a quadratic equation is not in factored form, we can put in in standard form and use the Quadratic Equation to solve for the roots. It looks complicated, but it’s not too bad, once you get the hang of it. The nice thing about the Quadratic Formula is that it can be used with any quadratic, even those without actual $ x$-intercepts, in which case they will be imaginary (see examples here in the Imaginary (Non-Real) and Complex Numbers section).
The $ \pm$ means that, since we typically have two solutions, unless the vertex is on the $ x$-axis, in which case we have one solution. We need to find one solution using a “$ +$” in that part of the equation, and one using a “$ -$” there. Here’s an example:
The Quadratic Formula is a very important thing to remember; you may even learn a song to help you remember it! You may sing it to “Pop Goes the Weasel”: “$ x$ equals negative $ b$, plus or minus the square root, of $ b$ squared minus $ 4ac$, all over $ 2a$”.
Now let’s get the roots by using the complete quadratic equation for our bouquet-throwing equation.
These roots are a little more complicated, since they aren’t rational; which means they actually go on and on “forever”. We’ll find the answers to the thousandths place (3 decimal places after the decimal point). This is one we couldn’t have factored!
Which of the roots do we want when we are getting the time the bouquet hits the ground? Since time can’t be negative, we want the positive one! Many times in math application, we need to “throw away” some or all of the answers.
The bouquet hit the ground approximately 1.464 seconds after the bride threw it up in the air. (Note that this is a time, not a distance; sometimes parabolas represent the distance on the $ x$-axis and the height on the $ y$-axis, and the shapes are similar.)
Note that, from above, $ \displaystyle \text{ }\frac{{5-3\sqrt{5}}}{8}\text{ and }\frac{{5+3\sqrt{5}}}{8}$ (also written as $ \displaystyle \frac{5}{8}-\frac{{3\sqrt{5}}}{8}\text{ and }\frac{5}{8}+\frac{{3\sqrt{5}}}{8}$) are called conjugates or radical conjugates, since they are identical, except for the change in signs in the middle of the real terms and the irrational terms. When we have an irrational value for roots, the conjugate is also a root. As other examples, if $ \displaystyle 3+\sqrt{{17}}$ is a root, then $ \displaystyle 3-\sqrt{{17}}$ is also a root; if $ \sqrt{2}$ is a root, then $ -\sqrt{2}$ is also a root.
Let’s ask one more question, and you can see the result on the graph below. How high will be bouquet be after 1 second, and is it on its way up, or its way down?
At one second, we can plug in “1” for the $ t$, since the $ t$-axis (this is usually what we call the $ x$-axis) is time. We have:
$ \begin{array}{c}f(t)=-16{{t}^{2}}+20t+5\,\\f(1)=-16{{(1)}^{2}}+20(1)+5=-16+20+5=9\end{array}$
This means at 1 second, the bouquet is 9 feet off the ground. Look at the graph below; at 1 second, it’s after the time when it was at its highest, so the bouquet is on its way down. We can also see that the time of 1 second is later than the time of the highest point (vertex) at .625 seconds to know that it’s on its way down.
Using the Discriminant to Determine Types of Solutions
It turns out that what’s under the radicand in the Quadratic Equation, which is $ \displaystyle {{b}^{2}}-4ac$ for the equation $ a{{x}^{2}}+bx+c=0$, happens to be very important; so important, that’s it’s given the name “the discriminant.” It helps us to discriminate between the type and number of $ \boldsymbol{x}$–intercepts, or “real roots” that we’ll get from that equation.
Note that we use the word “real” to indicate that the root isn’t imaginary. In fact, only real roots have $ \boldsymbol{x}$–intercepts, as shown:
Using the Graphing Calculator to Find the Vertex and Solve Quadratics
We can use the graphing calculator to graph the quadratic, and also find the vertex (which is the minimum or maximum), the roots, $ y$-intercept, and any values on the graph. Some teachers will let you use it during tests! If you need a quick introduction to the graphing calculator, read the Introduction to the Graphing Calculator section first. (I’m using the TI-84 Plus CE calculator.)
NOTE: If at any time the graphing calculator gets “stuck” (blinking line in top right of screen), simply turn the calculator OFF (2nd ON), and then back ON again.
To find the Vertex of a quadratic equation using a graphing calculator:
You can find the $ \boldsymbol{x}$–intercepts, or roots of a quadratic (or any polynomial) equation as follows. Note that when the parabola doesn’t cross the $ x$-axis, and thus there are no real solutions, you’ll get an error in your calculator when you try to find the zeros.
We can also use the calculator to get any $ y$-value for $ x$, the $ \boldsymbol{y}$-intercept, and see a table of the $ x,y$-values:
There’s one more really neat thing we can do with the calculator; we can check the answers we get from the Quadratic Formula – even if they are quite complicated. To do this, we’ll store the answers we get in a variable $ x$ in the calculator, and then type in the quadratic to see if we used the Quadratic Formula correctly (the quadratic should be 0).
Quadratic Transformations
We will learn about transforming parent functions here in the Parent Functions and Transformations section, but transforming quadratics isn’t that difficult for horizontal parabolas.
The parent function for quadratics is $ \boldsymbol{{x}^{2}}$ with a vertex at $ (0,0)$. We can see most easily see a quadratic transformation by putting the quadratic into vertex form, and plotting the new vertex. When the vertex is other than $ (0, 0)$, we have a horizontal and/or vertical shift of the parent function. When there is no vertical or horizontal stretch, we go over 1 (on each side of the vertex, since the function is symmetrical) and either up or down 1 to get the points close to the vertex. When the parabola is “upside down”, there is a negative before the term with the $ {x}^{2}$.
Thus, for the transformed function $ y=a{{\left( {x-h} \right)}^{2}}+k$, $ (h,k)$ is the vertex, and “$ a$” is a vertical stretch. (If “$ a$” isn’t 1, go over 1 or back 1, but then go up or down the value of “$ a$”). If the “$ a$” is negative, do the same with a flipped (facing down) graph. Then you can plug in other values of “$ x$” close to the axis of symmetry, to get the new values for “$ y$”. (For a horizontal stretch, such as $ y={{\left( {b\left( {x-h} \right)} \right)}^{2}}+k$, go up 1 or down 1 from the vertex, and then go over and back the value of “$ \displaystyle \frac{1}{b}$”).
Here are some examples; we can also do these transformations with t-charts like we do here in the Parent Functions and Transformations section:
Here are other types of problems you may see:
Quadratic Regression
We learned about regression here in the Scatterplots, Correlation, and Regression section; let’s find a Quadratic Regression.
Model the following data set using the graphing calculator.
| $ x$ | -3 | 1 | 2 | 4 | 6 | 8 | 10 |
| $ y$ | 35 | 2 | 9 | 41 | 100 | 180 | 290 |
Enter the data and perform regression in the calculator.
Solving Quadratics – and there’s more!
Again, there are several ways to solve quadratics, or find the solution (also known as the $ x$-intercepts, zeros, roots, or values) of a quadratic equation. We just learned about two of them:
- Quadratic Formula
- Graphing the Quadratic (this will also apply to general polynomials, as we’ll see later)
We will talk about two more ways in the next section, Solving Quadratics by Factoring and Completing the Square:
- Factoring and setting factors to 0. However, not every quadratic can be factored.
- Completing the square and taking the square root of each side. We’ll also use this to go from Standard Form to Vertex Form.
Learn these rules, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
On to Solving Quadratics by Factoring and Completing the Square – you are ready!