 Quadratics and the Parabola
 Graphing Quadratics (Parabolas)
 Standard Form to Vertex Form
 Solving for Roots with Quadratics
 Quadratic Transformations
 More Practice
Quadratics and the Parabola
One special type of polynomial equation that you’ll work with a lot is a Quadratic equation.
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).
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.
The main thing we have to remember about quadratics is that they either go up or back down (like our bouquet), or they go down and then back up. (Actually, they can go sideways, too, but that discussion is in a more advanced section). So they either have a maximum point, or a minimum point; whatever this point is, it’s called the vertex.
Let’s look at the basic equations for parabolas. As we said before, they are some variation of \(y={{x}^{2}}\) (they can also be \(x={{y}^{2}}\), which are the “sideways” parabolas and are not functions, since they don’t meet the vertical line test). But the quadratic equations we’ll be dealing with now will either go straight up or straight down.
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?
Equation 
Parabola Form 
Parabola Characteristics from this Form 
\(y=a{{x}^{2}}+bx+c\) 
Standard Form 
\(c\) is the \(y\)intercept 
\(\begin{array}{l}y=a{{\left( {xh} \right)}^{2}}+k\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{or }\\yk=a{{\left( {xh} \right)}^{2}}\end{array}\) 
Vertex Form 
\((h,k)\) is the vertex 
\(y=a\left( {x{{x}_{1}}} \right)\left( {x{{x}_{2}}} \right)\) 
Factored Form (also called Intercept Form) 
\({{x}_{1}},{{x}_{2}}\) (at the points \(\left( {{{x}_{1}},0} \right)\) and \(\left( {{{x}_{2}},0} \right)\)) are the \(x\)intercepts or zeros (if the parabola touches/cross the \(x\)axis).
Zeros are also called values, solutions, or roots. 
One thing that’s very important in looking at all these forms is when the coefficient of \(\boldsymbol{{x}^{2}}\)^{ }(the “\(a\)”) is positive, the shape of the graph is like a “\(\bigcup \)”, and has a minimum for the vertex. If the “\(a\)” is negative, the shape is like a mountain, or a “\(\bigcap \)” and the graph has a maximum for the vertex.
Another thing to know about parabolas is that the number without an “\(x\)” (the “\(c\)”, or constant) is where the graph goes through the \(y\) axis; this is the \(y\)intercept. We’ll talk about how to get the \(x\)intercepts later.
Here are some examples of parabolas in standard form, with their characteristics:
Equation 
Characteristics 
\(y={{x}^{2}}\) 
\(\bigcup \) shape, \(y\)intercept is \((0,0)\) 
\(y=4{{x}^{2}}+6x+3\) 
\(\bigcup \) shape, \(y\)intercept is \((0,3)\) 
\(y={{x}^{2}}4\) 
\(\bigcap \) shape, \(y\)intercept is \((0,–4)\) 
\(y=10{{x}^{2}}+6x\) 
\(\bigcap \) shape, \(y\)intercept is \((0,0)\) 
Graphing Quadratics (Parabolas)
Before we talk specifically about the Vertex and Factored forms, let’s first create a tchart so we can graph the simplest form of a parabola, which is \(y={{x}^{2}}\).
Note that parabolas have symmetry; it is a mirror image of itself across the vertical line (called the line of symmetry, or LOS, or sometimes called the axis of symmetry) that contains its vertex. 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?
Note that since this equation (\(y={{x}^{2}}\)) is the simplest form of a parabola, it is called a parent function. We will learn about other parent functions later.
You’ll probably be asked to get the Domain (all the possible \(x\) values) and Range (all the possible \(y\) values) when you graph the parabolas. Remember to use your pencil to check the domain and range like we did here in the Algebraic Functions section.
Note that the Domain is \(\left( {\infty ,\infty } \right)\), or all real numbers, since these types of parabolas go up and out to the sides forever. 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.
Here’s our first parabola:
tchart  Graph  
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left[ {0,\infty } \right)\) 
Graphing with Vertex Form
Now 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:
Generic Vertex Form 
Vertex 
Example 
Vertex 
\(y=a{{\left( {xh} \right)}^{2}}+k\)  \(\left( {h,\,k} \right)\)  \(y=3{{\left( {x+2} \right)}^{2}}1\)  \(\left( {2,\,1} \right)\) 
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 \(y=3{{(x+2)}^{2}}1\). Notice that we don’t know right away what the \(y\)intercept is, but since we get the \(y\)intercept when \(x=0\), we can see that the \(y\)intercept is \(y=3{{(0+2)}^{2}}1=3{{(2)}^{2}}1=121=11\); the point is \((0,11)\).
We can graph the vertex and the \(y\)intercept, and then do a quick tchart to find some of the other values. Notice how 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.
Note that 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.
tchart  Graph  
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left[ {1,\infty } \right)\) 
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 \((0,11)\) would be \((4,11)\). 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 \(y=2{{\left( \frac{1}{2}x+2 \right)}^{2}}3\), and we can do some algebra to change it into \(y=2{{\left( \frac{1}{2}\left( x+4 \right) \right)}^{2}}3=2{{\left( \frac{1}{2} \right)}^{2}}{{\left( x+4 \right)}^{2}}3=\frac{1}{2}{{\left( x+4 \right)}^{2}}3\), so the vertex is \((4,3)\).
Graphing with Standard Form
Other times, when we have the equation in standard form, one of the ways we can get the vertex and \(y\)intercept is using the “formulas” below. (The other way is by Completing the Square, which we learn how to do here in the Solving Quadratics by Factoring and Completing the Square section.)
Generic Standard Form 
Vertex 
\(\boldsymbol{y}\)intercept 
\(y=a{{x}^{2}}+bx+c\) 
\(\displaystyle \left( {\frac{b}{{2a}},\,\,\,\,f\left( {\frac{b}{{2a}}} \right)} \right)\) or \(\displaystyle \left( {\frac{b}{{2a}},\,\,\text{plug in}\,\,\frac{b}{{2a}}\,\,\,\text{into the }x\text{ to get the }y\text{ }} \right)\) 
\(\left( {0,\,\,c} \right)\) 
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.
Then, to get the vertex, we can plug in the line of symmetry \(x\) point in the original equation to get the \(y\).
Factored Form Equation  How to Graph Example  Other Notes 
\(\displaystyle \begin{array}{c}\text{Generic:}\\y=a\left( {x{{x}_{1}}} \right)\left( {x{{x}_{2}}} \right)\\\\\\\text{Example:}\\\color{#800000}{{y=3\left( {x+4} \right)\left( {x2} \right)}}\end{array}\) 
We know that –4 and 2 are the \(x\)intercepts, from the Factored Form (Intercept Form) equation.
Now we can get the Line of Symmetry or LOS of the quadratic by taking the average (middle) of the two \(x\)intercepts; this would be \(\displaystyle x=\frac{{4+2}}{2}=1\).
We can get the vertex coordinates of the parabola by putting in –1 for \(x\) to get \(y\): \(y=3\left( {1+4} \right)\left( {12} \right)=27\).
We have vertex \(\left( {1,27} \right)\) and two points \(\left( {4,0} \right)\) and \(\left( {2,0} \right)\), so we can graph the parabola! 
Remember that when we have a factor of \(x\) alone, such as in \(y=3x\left( {x2} \right)\), the \(x\) is the same as \((x0)\), so this \(x\)intercept is \(\left( {0,0} \right)\). Therefore, in the example of \(y=3x\left( {x2} \right)\), the two \(x\)intercepts would be 0 and 2.
If we just have a quadratic in the form \(y=a{{x}^{2}}\), the only \(x\)intercept is \(\left( {0,0} \right)\), and the LOS is \(x=0\).
We can easily get more points on the graph by picking a random \(x\) value and then plugging in to find the corresponding \(y\) value. 
Standard Form to Vertex Form
Let’s look at some examples where we get the vertex from a standard form equation, since you’ll probably have to do this a lot. Remember that when the coefficient of \({{x}^{2}}\) is positive, the parabola faces up, and when the coefficient is negative, it faces down.
Quadratic Equation 
\(x\) value of Vertex 
\(y\) value of Vertex (plug \(x\) value into equation to get \(y\) value) 
Vertex/ Line of Symmetry 
Domain and Range 
\(y={{x}^{2}}\)
\(a=1,\,\,b=0,\,\,c=0\) Direction: Up 
\(\displaystyle \frac{b}{{2a}}=\frac{0}{{2\left( 1 \right)}}=0\)  \(y={{0}^{2}}=0\)  \(\begin{array}{c}\left( {0,\,0} \right)\\x=0\end{array}\) 
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left[ {0,\infty } \right)\) 
\(y={{x}^{2}}\)
\(a=1,\,\,b=0,\,\,c=0\) Direction: Down 
\(\displaystyle \frac{b}{{2a}}=\frac{0}{{2\left( {1} \right)}}=0\) 
\(y={{0}^{2}}=0\) 
\(\begin{array}{c}\left( {0,\,0} \right)\\x=0\end{array}\) 
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left( {\infty ,\,\,0} \right]\) 
\(y=5{{x}^{2}}20\)
\(a=5,\,\,b=0,\,\,c=20\) Direction: Up 
\(\displaystyle \frac{b}{{2a}}=\frac{0}{{2\left( 5 \right)}}=0\) 
\(y=5{{(0)}^{2}}20=20\) 
\(\begin{array}{c}\left( {0,\,20} \right)\\x=0\end{array}\) 
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left[ {20,\infty } \right)\) 
\(y=2{{x}^{2}}+6x\)
\(a=2,\,\,b=6,\,\,c=0\) Direction: Down 
\(\displaystyle \frac{b}{{2a}}=\frac{6}{{2\left( {2} \right)}}=\frac{3}{2}\) 
\(\displaystyle \begin{array}{l}y=2{{\left( {\frac{3}{2}} \right)}^{2}}+6\left( {\frac{3}{2}} \right)\\\,\,\,\,\,=\frac{9}{2}+9=\frac{9}{2}\end{array}\) 
\(\displaystyle \left( {\frac{3}{2},\,\,\frac{9}{2}} \right)\) \(\displaystyle x=\frac{3}{2}\) 
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\displaystyle \left( {\infty ,\,\,\frac{9}{2}} \right]\) 
\(y={{x}^{2}}+4x+3\)
\(a=1,\,\,b=4,\,\,c=3\) Direction: Down 
\(\displaystyle \frac{b}{{2a}}=\frac{4}{{2}}=2\) 
\(\begin{array}{l}y={{\left( 2 \right)}^{2}}+4\left( 2 \right)+3\\\,\,\,\,\,=4+8+3=7\end{array}\) 
\(\begin{array}{c}\left( {2,\,\,7} \right)\\x=2\end{array}\) 
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left( {\infty ,\,\,7} \right]\) 
Let’s graph the last equation above \(y={{x}^{2}}+4x+3\). Remember that the \(y\)intercept is when \(x=0\), so the \(y\)intercept is \((0,3)\). And we know that the vertex is \((2,7)\).
Some other points:
tchart 
Graph 

Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left( {\infty ,7} \right]\) 
Now let’s use quadratics to model the flight of our bride’s bouquet.
Let’s have 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 the function notation? If not go back and review.
(Just for some history and a tiny bit of physics: The Italian scientist Galileo discovered the parabolic trajectory of projectiles in the early 17^{th} century. Neglecting air friction, the \(16{{t}^{2}}\) represents the gravity of the bouquet in feet per second per second (actually the 16 is half the force of gravity). The 5 feet represents the height at which the bouquet is thrown at the beginning, and the 20 in the 20t represents the initial velocity (feet per second) vertically upward, or how fast the bride throws the bouquet up in the air.)
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 (sorry, girls…).
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 “\(\displaystyle \frac{b}{{2a}}\)” trick to obtain the vertex (highest point), and \(y\)intercept to graph our equation:
Standard Form 
Vertex 
\(y\)intercept 

Generic 
\(f\left( x \right)=a{{x}^{2}}+bx+c\) 
\(\displaystyle \left( {\frac{b}{{2a}},\,\,\,\,f\left( {\frac{b}{{2a}}} \right)} \right)\) or (\(\displaystyle \frac{b}{{2a}}\) is the \(x\), plug \(\displaystyle \frac{b}{{2a}}\) into the \(x\) to get the \(y\)) 
\(\left( {0,\,\,c} \right)\) 
Our Equation 
\(f(t)=16{{t}^{2}}+20t+5\) 
\(\displaystyle \frac{b}{{2a}}=\frac{{20}}{{2(16)}}=\frac{{20}}{{32}}=\frac{5}{8}=.625\)
\(\begin{align}y&=16{{(.625)}^{2}}+20(.625)+5\\\,&=6.25+12.5+5\\\,\,\,&=11.25\end{align}\)
Vertex: \(\left( {.625,\,11.25} \right)\) 
\(\left( {0,\,\,5} \right)\) 
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 \(\boldsymbol{x}\)–intercept; these are called “roots” of the quadratic equation. This is actually how we “solve” quadratic equations. Sometime quadratics never hit the \(\boldsymbol{x}\)axis; these quadratics still have roots, but they are “imaginary”, and we’ll address them later in the Imaginary (NonReal) and Complex Numbers section.
Quadratic Formula for Solving Roots
We may want to find out when the bouquet hits the ground. Again, this is the \(\boldsymbol{x}\)–intercept, since it’s where the graph intercepts the \(x\)axis, and it’s where \(y=0\). It’s also called a root of, a value of, zero of, a solution of and solving a quadratic equation.
This is because setting a quadratic equation to 0 is the most direct way to solve it. This is a very important concept. So (in most cases, as we see) if we’re given an equation with an \(\boldsymbol{{x}^{2}}\) in it, we should move everything to one side, and have 0 on the other side.
In quadratic equations, there are usually 2 \(x\)intercepts or roots, but there may be only 1, or none.
The \(\boldsymbol{x}\)–intercept is when \(y=0\), so we need to solve the following equation for \(t\):
\(0=16{{t}^{2}}+20t+5\)
This is a little difficult to get, as compared to a linear function. What do we do with the \({{t}^{2}}\)? We can’t solve this like we did before with linear equations.
Someone came up with this really cool way to find the \(x\)intercepts; it’s called the Quadratic Formula. Now it looks complicated, but it’s not too bad, once you get the hang of it. Again, this solves for the \(\boldsymbol{x}\)–intercepts, which are also called zeros, values, solutions, and roots. And if there are no \(x\)intercepts, we can still get the roots for a quadratic, but they will be “imaginary” (see examples here in the Imaginary (NonReal) and Complex Numbers section).
The \(\pm\) means that, since we typically have two solutions, we need to find one solution using a “\(+\)” in that part of the equation, and one using a “\(\)” there.
Let’s do an easier (than our bouquet) example first:
Standard Form/ Quadratic Formula 
Example of Solving for \(\boldsymbol{x}\)intercept 
Graph 
\(y=a{{x}^{2}}+bx+c\)
\(\displaystyle x=\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}\)
We are actually solving the equation:
\(0=a{{x}^{2}}+bx+c\)

\(y={{x}^{2}}2x8\) \(a=1,\,\,\,b=2,\,\,\,c=8\)
\(\displaystyle \begin{align}\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}&=\frac{{\left( {2} \right)\,\,\pm \,\,\sqrt{{{{{\left( {2} \right)}}^{2}}4\left( 1 \right)\left( {8} \right)}}}}{{2\left( 1 \right)}}\\&=\frac{{2\,\,+\,\,\sqrt{{4\left( {32} \right)}}}}{2}\text{ }\,\,\text{ }\\&\text{and }\,\frac{{2\,\,\,\,\sqrt{{4\left( {32} \right)}}}}{2}\\&=\frac{{2\,\,\,\,\sqrt{{36}}}}{2}\text{ and }\,\frac{{2\,\,+\,\,\sqrt{{36}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\end{align}\)
\(\displaystyle x\,=\,2\text{ and 4 }\) 
The \(x\)intercept, or roots are \((2,0)\) and \((4,0)\).
See how this is when \(y=0\)? 
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 bouquetthrowing equation. We see that it has 2 roots from evaluating our discriminant, but since the discriminant isn’t a perfect square, these roots aren’t rational, which means they have funny decimals in them (that go on forever).
Let’s just work with decimals to the thousandths place (3 decimal places after the decimal point).
Bouquet Equation 
Discriminant (see below) 
Roots 
\(y=16{{t}^{2}}+20t+5\) 
\(\begin{align}{{b}^{2}}4ac&={{\left( {20} \right)}^{2}}4\left( {16} \right)\left( 5 \right)\\\,\,\,\,\,&=400\left( {320} \right)\\\,\,\,\,\,&=720\end{align}\)
\(720 > 0\)
2 roots! 
\(\displaystyle \begin{align}\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}&=\frac{{20\,\,\pm \,\,\sqrt{{720}}}}{{2\left( {16} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac{{20\,\,+\,\,26.833}}{{32}}\text{ }\text{ and }\text{ }\frac{{20\,\,\,\,26.833}}{{32}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\,\,\,\,.214\text{ and 1}\text{.464 }\end{align}\)
The \(x\)intercepts or roots are \(\displaystyle \left( {.214, 0} \right)\) and \(\displaystyle \left( {1.464, 0} \right)\).
Exact answers are \(\displaystyle \left( {\frac{{53\sqrt{5}}}{{8}},0} \right)\) and \(\displaystyle \left( {\frac{{5+3\sqrt{5}}}{{8}},0} \right)\), since \(\require{cancel} \displaystyle \frac{{20\,\pm \,\sqrt{{720}}}}{{2\left( {16} \right)}}=\frac{{20\,\pm \,\sqrt{{\left( {144} \right)\left( 5 \right)}}}}{{32}}=\frac{{{{{\cancel{{20}}}}^{5}}\pm \,{{{\cancel{{12}}}}^{3}}\sqrt{5}}}{{{{{\cancel{{32}}}}^{8}}}}\). 
Note that \(\displaystyle \frac{{5+3\sqrt{5}}}{{8}}\text{ }\,\,\text{and }\frac{{53\sqrt{5}}}{{8}}\) (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.
Note again that we just “solved” the equation; to solve a quadratic equation, we find the \(x\)intercepts (also called roots, solutions, zeros), or where the graph hits the \(x\)axis (this is when \(y=0\)). Even if we have no \(x\)intercepts, we can still solve the quadratics to get the values, solutions, or roots of the equation; these will then be “imaginary”. We can also get the roots of quadratics (sometimes) by what we call factoring the quadratic; we’ll do this in the next section Solving Quadratics by Factoring and Completing the Square.
But looking back at the graph, we see that if we want to know when the bouquet hits the ground, we need the second answer – the positive one. We need to “throw away” the first answer, since it really happened before we threw the bouquet (and there really isn’t such thing as a negative time).
The bouquet hit the ground 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.)
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. So 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
Now, what’s under the radicand in the Quadratic Equation, which is \(\displaystyle {{b}^{2}}4ac\), happens to be very important; so important, that’s it’s given the name “the discriminant.” It’s given this name since 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:
Discriminant  Number of roots  Example  Graph 
\(\displaystyle {{b}^{2}}4ac=0\) 
1 real root
Touches \(x\)axis once 
\(y={{x}^{2}}6x+9\)
\(\displaystyle \begin{array}{c}{{b}^{2}}4ac=\\{{\left( {6} \right)}^{2}}4\left( 1 \right)\left( 9 \right)=\\3636=\,\,0\end{array}\) 

\({{b}^{2}}4ac>0\) 
2 real roots
Touches \(x\)axis twice 
\(\displaystyle y={{x}^{2}}2x+2\)
\(\begin{array}{c}{{b}^{2}}4ac=\\{{\left( {2} \right)}^{2}}4\left( {1} \right)\left( 2 \right))=\\4+8=12\end{array}\) 

\({{b}^{2}}4ac<0\) 
No real roots
Doesn’t touch \(x\)axis: no \(x\)intercepts (Imaginary roots) 
\(y={{x}^{2}}2x+2\)
\(\begin{array}{c}{{b}^{2}}4ac=\\{{\left( {2} \right)}^{2}}4\left( 1 \right)\left( 2 \right)=\\48=4\end{array}\) 

And one more thing that’s interesting:
If \(\displaystyle {{b}^{2}}4ac=\) a perfect square, such as \((0,1,4,9,16,25,36,…)\) 
2 real rational (“easy”) roots (1 root if discriminant \(=0\))
(We’ll see later that these quadratics can be factored) 
\(y={{x}^{2}}x6\)
\(\begin{array}{c}{{b}^{2}}4ac=\\{{\left( {1} \right)}^{2}}4\left( 1 \right)\left( {6} \right)=\\1+24=25\end{array}\)


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, 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.
NOTE: If at any time the graphing calculator gets “stuck” (blinking line in top right of screen), simply turn the calculator OFF (2^{nd} ON), and then back ON again.
To find the Vertex of a quadratic equation using a graphing calculator:
Calculator Instructions to Find Vertex of Quadratics  Screens 
If Plot1 (or Plot2 or Plot3) is turned on (if it is highlighted or dark), turn it off by moving cursor up to it and hitting ENTER. The highlighting should go away.
(You can always clear the calculator completely by hitting “2^{nd} +” (MEM) and then 7 (Reset), then 1 (All RAM), then 2 (Reset). Do this with caution; it will wipe out all your data and programs in your calculator!)
Push “Y =” and enter the equation \(f(t)=16{{t}^{2}}+20t+5\) in “Y_{1} =”. (You may need to press CLEAR after “Y_{1} =” first to clear out older equations). For the \({{t}^{2}}\) part, either use the x^{2} button, or use x^2. Unless you have a very old version of the calculator Operating System, after the x^2, you have to use the right cursor button to move the cursor back down after putting in the exponent.
Push GRAPH. You may need to hit “ZOOM 6” (ZoomStandard) and “ZOOM 0” (ZoomFit) and “ZOOM 3” (Zoom Out) ENTER (maybe more than once) to make sure you see the vertex in the graph.

You can also use the WINDOW button to change the minimum and maximum values of your \(x\) and \(y\) values; you’ll just want to play around with the values until you can see the vertex; in this case the maximum. 
To get the vertex, push “2^{nd} TRACE” (CALC), and then either push 3 for MINIMUM or 4 for MAXIMUM. In our case, we’ll push 4, or move the cursor down to MAXIMUM.
The calculator will say “Left Bound?” and then, using the cursors, move the cursor anywhere to the left of the maximum point and hit ENTER. Ignore the \(x\) and \(y\) values.
When the calculator says “Right Bound?”, move the cursor anywhere to the right of the maximum point and hit ENTER. Ignore the \(x\) and \(y\) values.
The calculator will say “Guess?”. Hit ENTER once more, and you have your maximum point, or vertex. 
Yeah! We got the same vertex that we did when we used the Quadratic Formula!

To find the \(\boldsymbol{x}\)–intercepts, or roots of a quadratic (or any polynomial) equation:
Graphing Calculator Instructions for Roots of Quadratics  Screens 
To get the roots, push “2^{nd} TRACE” (CALC), and then push 2 for ZERO (or move cursor down to ZERO).
The calculator will then say “Left Bound?” Using the cursors, move the cursor anywhere to the left of the first zero (where the graph hits the \(x\)axis) and hit ENTER.
When the calculator says “Right Bound?” move the cursor anywhere to the right of the first zero and hit ENTER.
The calculator will say “Guess?”. Hit ENTER once more, and you have your leftmost zero. Do the same for the other zero, if there is one. Note: You can use the TRACE button to move the cursor closer to one of the zeros first if you want to. 
For example, to get the rightmost (nonnegative) zero:

Note that you can also find the roots by setting “Y_{2} =” to 0 and use the Intercept function to find the roots.
You may have to use the ZOOM and/or WINDOWS to make sure you see the point of intersection you want (I like to use ZOOM 6, ZOOM 0, and then ZOOM 3 ENTER, ZOOM 3 ENTER, and so on).
You may also use TRACE to move the cursor close to the point of intersection you want, if the cursor is closer to the other root.
To get the point(s) of intersection, push “2^{nd} TRACE” (CALC), and then either push 5, or move cursor down to intersect. You should see “First curve?” at the bottom. Then push ENTER, ENTER, ENTER. 

Note that if there are no real solutions (the parabola doesn’t cross the \(x\)axis), 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:
Graphing Calculator Instructions to get Values  Screen 
To put in an \(x\) value to get a \(y\) value, push “2^{nd} TRACE” (CALC), and then push 1 for VALUE (or just hit ENTER).
You will see an “X =” on the lower left of the screen. Just type in the value, such as “1”. Hit ENTER.
You’ll see the \(y\) value on the bottom. 

To get the \(y\)–intercept, put in 0 for \(x\) this same way.  
You can also use the TABLE function to see different values for \(x\) and \(y\).
You can use “2^{nd} WINDOW” (TBLSET) to set where the table starts and what the \(x\) increments are. We start at \(x=0\) and use increments of .25 in this example. 

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 got the right answer (the quadratic should be 0) if we used the Quadratic Formula correctly:
Quadratic Formula Problem to Check  Instructions  Screen 
\(y=16{{t}^{2}}+20t+5\)
\(\displaystyle \frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}=\frac{{20\,\,\pm \,\,\sqrt{{720}}}}{{2\left( {16} \right)}}\) \(\require {cancel} \displaystyle =\frac{{{{{\cancel{{20}}}}^{5}}\pm \,\,{{{\cancel{{12}}}}^{3}}\sqrt{5}}}{{{{{\cancel{{32}}}}^{8}}}}\) \(\displaystyle =\,\,\frac{{5+3\sqrt{5}}}{{8}}\text{ and }\frac{{53\sqrt{5}}}{{8}}\text{ }\)

Type in one of the solutions, including parentheses around the numerator, and then hit
STO> X,T, ,t as shown:
Do the same for the other solution; you can use the buttons to go back to what you’ve already typed, and edit, to save time. 
Then type in the quadratic, using thebutton for \(x\). It will use the value you have in \(x\), so if you solved the quadratic correctly, you should get 0:

Quadratic Transformations
We will learn about Parent Functions and Transformations here, but transforming quadratics isn’t that difficult for horizontal parabolas.
The parent function for quadratics is \(\boldsymbol{{x}^{2}}\) with a vertex at \((0,0)\). Find the new vertex by putting the quadratic into vertex form, and plot the new vertex. When the vertex is other than \((0,0)\), we have a horizontal and/or vertical shift of the parent function.
For 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.
In the transformed function \(y=a{{\left( {xh} \right)}^{2}}+k\), “\(a\)” is a vertical stretch, so 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 \(y={{\left( {b\left( {xh} \right)} \right)}^{2}}+k\), go up 1 or down 1, but then go over or back the value of “\(\displaystyle \frac{1}{b}\)”).
Here are some examples:
Original Function  Shift and Vertical Stretch  Shift and Horizontal Stretch 
\(y={{x}^{2}}\)
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left( {\infty ,\infty } \right)\) 
\(\displaystyle y=5{{\left( {x3} \right)}^{2}}2\)
New vertex is \(\left( {3,2} \right)\) (right 3 and down 2). Flip graph because of – sign. Vertical stretch is 5: from vertex, go back 1, down 5; over 1, down 5:
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left( {\infty ,2} \right]\) 
\(\displaystyle y={{\left( {2\left( {x+2} \right)} \right)}^{2}}5\)
New vertex is \(\left( {2,5} \right)\) (left 2 and down 5). Horizontal stretch is \(\frac{1}{2}\)(reciprocal of 2, since it’s horizontal): from vertex go up 1, back \(\frac{1}{2}\); up 1, over \(\frac{1}{2}\):
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left[ {5,\infty } \right)\) 
Here are other types of problems you may see:
Quadratic Transformation  Solution 
Write an equation of a quadratic function that has been horizontally stretched by a factor of 4, and has a vertex of \(\left( {4,1} \right)\).  Since we have a horizontal stretch by a factor of 4, we know the equation is in the form \(y={{\left( {b\left( {xh} \right)} \right)}^{2}}+k\). \(\displaystyle \frac{1}{b}\) is the horizontal stretch (“opposite math” for horizontal transformations), and \(\left( {h,k} \right)\) is the vertex.
Thus, we have \(\displaystyle y={{\left( {\frac{1}{4}\left( {x4} \right)} \right)}^{2}}1\). 
Write an equation for the following quadratic:

Since we have a flipped parabola with a vertical stretch, we know the equation is in the form \(y=a{{\left( {xh} \right)}^{2}}+k\), where \(a\) is the vertical stretch, and \(\left( {h,k} \right)\) is the vertex.
The vertex is \(\left( {4,2} \right)\), and it looks like we have a vertical stretch of 3 (over 1, down 3, instead of over 1, down 1).
Thus, we have \(y=3{{\left( {x4} \right)}^{2}}+2\). 
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 (not all trinomials are factorable)
 Completing the Square (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!