Authors:
F12: Ryan Garman, Michael Bartus, Brad Mesclier
W11: Colleen Condra, Amy Pickell, Josh Swackhammer, Abby Quisenberry
F11: Matt York, Joe Freiman

Quadratic Functions are very applicable to every day life. The best example of a quadratic function is the path of a ball being thrown into the air. Below is a diagram demonstrating this motion.

Picture created by Amy Pickell

The arc as a whole can be plotted on a coordinate plane in the form of a quadratic graph. From this graph we can determine the equation of the parabola and the height of the ball. Using this equation we can determine the velocity and acceleration of the ball.
We will walk step-by-step through this process later in this section.

The history of quadratics goes way back to 2000 BC when the Babylonians solved quadratics in radicals. The idea of quadratics has greatly developed since then. Chinese and Babylonian mathematicians solved quadratic equations by completing the square. When they solved quadratics they would focus on finding a length, and the length that was found was the root of the quadratic equation. The first steps towards finding a general formula to solve quadratic equations trace back around 300 BC, mathematicians Pythagoras and Euclid found a method to solve the quadratic equation by focusing strictly on geometry. Pythagoras noted that the ratios found a general procedure to solve the quadratic equation. During Euclid's time, there didn't exist the same notation for numbers and formulas that exist today, and because of that, it was not possible to compute the square root of any number by hand. In 800 AD, al-Khwarizmi, a Persian mathematician, created six new equations involving roots, squares of roots, and numbers which he used to solve quadratic equations, and derived the quadratic equation. The algebra used by him was entirely rhetorical, and he rejected negative solutions. These discoveries led Abraham bar Hiyya Ha-Nasi to publish a book with the complete solution of the quadratic formula. In the year 1545, Gerolamo Cardano, a mathematician from Italy and also one of the best algebraists of his time, collected all the works related to equations. Later in the year 1594, the quadratic formula was fist obtained by Simon Stevin, who was born in Bruges (now known as Belgium). Everything we know today stems off of these earlier discoveries. The quadratic formula as know it today is

x1,2=(-b/2a) ± (1/2a)(b2-4ac)1/2,

and this gives the solution to a generic quadratic equation of the form:

In order for mathematics education to become more focused and coherent and to improve on overall mathematics achievement the states are implementing these common core state standards. The standards below are what Michigan requires of the mathematics curriculum:

Construct quadratic functions given a graph, a description of a relationship, or two input-output pairs.

Create equations and inequalities in one variable and use them to solve problems. Include equations from linear and quadratic functions, and simple rational and exponential functions.

Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x – p)2 = q that has the same solutions. Derive the quadratic formula from this form.

Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a± bi for real numbers a and b.

The NCTM standards that apply:

model real-world phenomena with a variety of functions;

represent and analyze relationships using tables, verbal rules, equations, and graphs;

translate among tabular, symbolic, and graphical representations of functions;

recognize that a variety of problem situations can be modeled by the same type of function;

analyze the effects of parameter changes on the graphs of functions;

understand operations on, and the general properties and behavior of, classes of functions.

The HSCE standards that apply:

Write the symbolic form and sketch the graph of a quadratic function given appropriate information.

Convert quadratic functions from standard to vertex form by completing the square.

Relate the number of real solutions of a quadratic equation to the graph of the associated quadratic function.

Express quadratic functions in vertex form to identify their maxima and minima, and in factored form to identify their zeros.

(GeoGebra file available at GeoGebra Tube) A quadratic function has a "U" shaped graph called a parabola that can either open up or down. When the parabola opens upward the coefficient in front of the quadratic term is positive whereas when the parabola opens downward the coefficient is negative. A quadratic function has a standard form of f(x)=ax^2 + bx + c where a cannot equal 0. As seen in the above Geogebra applet when you manipulate the values of a, b, and c your parabola seems to change. Changing the value of a causes your parabola to "shrink" or "stretch" in fact your parabola is actually going to infinity faster or slower than it was. Changing b causes a change in the downward slope of your parabola. Changing c changes where your parabola intersects the y-axis or it might also be called the height.

(Geogebra file hosted at Geogebra Tube.) The vertex form a parabola is f(x) = a(x - h)^2 + k where h and k are the (x,y) coordinates of the vertex of the parabola the sign of a determines which way the parabola will open. Observe this by changing the values of h, k, and a in the Geogebra applet above. When the quadratic equation is set equal to 0 and we solve for x we obtain the roots or zeros of the equation. These roots or zeros are when the values of x make f(x)=0 this is when our projectile is at a height of zero and crosses the x-axis.

How the Quadratic formula is related to a Quadratic Equation

In this section, we will show how closely connected the quadratic formula to a quadratic equation step-by-step with the picture below as a reference so that you may see what is being done.
+ For the first step, we will start with an arbitrary equation, where a is nonzero and a,b, and c, . which is stated in step one. Observe that our equation is a second degree polynomial equation. This means that we will have two solutions for our input (x). We will now subtract c from each side of the equation.
+ For step two, we will divide both sides of the equation by a.
+ From step three, we want to complete the square. To do that, we must add b2/4a2 to both sides of the equation.
+ From step four, we will complete the square for the left side of the equation.
+ From step five, we will combine the like-terms on the right side of the equation.
+ From step six, we will take square roots of both sides of the equation.
+ From step seven, we simplify the equation.
+ From step eight, we will subtract both sides of the equation by -b/2a
+ From step nine, we will combine like-terms.
+ From step ten, there is no need to continue on because we have found the quadratic formula!!!!!

Now that you've seen how the equation works, let's see some examples and other ways quadratic equations are useful!

Photo Credit to Brad Mesclier

Standard and Vertex Form of Quadratic Equations

The formula fora quadratic function is different from that of a linear function because it contains and x^2 term. Examples of quadratic functions include

f(x)= 2(x^2)-4x-11 g(x)=4-(x^2) h(x)=(1/3)(x^2)+(2/3)x+5 The general form of a quadratic function, also known as the standard form, is then as follows Let a,b, and c be constants with a /=/ 0. A function represented by f(x)=a(x^2)+bx+c is a quadratic function.
The domain of a quadratic function includes all real numbers and the leading coefficient of a quadratic function is a as listed above. For the equations listed above the value of a is as follows a = 2 a = -1 a = (1/3) When a quadratic function f is expressed as f(x)=a(x^2)+bx+c, the coordinates of the vertex are not apparent. However, if f is written as f(x)= a(x-h)^2 +k, then the vertex is located at (h,k). To justify that eh vertex is indeed (h,k), we can use an inequality of a. If a>0 in the form f(x)= a(x-h)^2 +k, then the term a(x-h)^2 is never negative and the minimum value of f(x) is k. This value occurs when x=h because f(h)=a(h-h)^2 +k = 0+k = k Thus the lowest point on the graph of f(x)=a(x-h)^2 +k with a>0 is (h,k) and because this graph is a parabola that opens upward, the vertex must be (h,k). You can prove this for a<0 in a similar fashion. So in summary the equation for the vertex form of a quadratic formula is as follows f(x)=a(x-h)^2+k

Examples:

The x-intercepts(roots/zeros) can be found using many different methods. One requires us to our quadratic function you then set the factors equal to zero and solve for your variable. Another has you use a formula also known as the quadratic formula where you can plug in the values for a, b, and c, given by the quadratic function in standard form (ax^2+bx+c). This formula is Also, graphing and looking for the x-intercepts and completing the square are useful methods to keep in mind.

This is an example for solving a quadratic equation for the x-intercepts(roots/zeros):

Begin with an equation. In this example we will use y = x^2 +4x+4.
For a quadratic it is important to factor the polynomial using the following steps.

1. Put y = 0 and insert this into the left side of the equation, such that
0 = x^2 +4x+4.
2. Determine how this can be factored. In this case we have to find two numbers that will equal 4 when multiplied, and also can be added together to get 4.
Options: (4 and 1), (2 and 2).
3. Once in factor form, we can set each factor equal to zero and solve for x, in this case
(x+2)(x+2) = 0
(x+2) = 0
(x+2) = 0
x = -2, -2
Therefore the x-intercepts, roots, or zeros are -2 and -2 in this case.

This is an example of solving using the quadratic formula:

We begin with the equation x^2 +4x + 4. For the sake of the example and simplicity, we will pretend it can't be factored. We see the quadratic forumula:
x=

We have 3 variables: a,b, and c. Our equation is in standard form: ax^2+bx+c = 0. In our equation, we can see that a=1, b= 4, and c= 4.
Now we have everything we need; from here on out, it's just a matter of plugging in and making calculations.

Start by entering in values:

(-4±√(4^2 - (4*1*4) ))/(2*1)

Our determinant (the square root portion of the forumula) = 0.
This gives us 2 roots. -2, and -2 (since -4±0 = -4, and -4/2 = -2)

This checks with how we solved above with factoring. We could also check by looking at a graph of our function and making sure it crosses the x-axis only at -2.

Extension question: Our only root is -2, how is this possible with a parabola? Based on our coefficients, is it pointing up or down? Does this mean that the point (0,-2) is a maximum point or minimum?

Since there is only one point crossing the x-axis, this point must be a the vertex. Looking at our coefficient of 1, we can conclude that our parabola opens up at the point (0,-2). This is a global minimum.

This is an example of solving a story problem using the quadratic formula:

You were out hiking behind you house in the hills of North Dakota when you lost your footing and fell down a steep hill. When you reached the bottom you realized you broke your leg and there is no way you can make it back home by yourself. Being an experienced hiker you brought your flare gun in case of this occurrence. You know that the flare must be above 200 feet in the air of someone back home to see it. Given the equation -16t+130t+1, given that t is in seconds, how long will the flare be in the air at 200 feet or higher?

Start by setting your equation equal to 200 feet:

200=-16t+130t+1

Subtract 200 feet from both sides:

0=-16t+130t-199

Now use the quadratic formula to solve for t:
(-130±√(130^2 - (4*16*-199) ))/(2*-16)

Simplify:

(-130±√(4164))/(-32)

Now solve:

(-130+64.53)/-32=2.05

(-130-64.53)/-32=6.08

Therefore your flare will be at 200 feet or higher from 2.05 seconds to 6.08 seconds. Now find the total time the flare will be 200 feet or higher.

6.08-2.05=4.03

Answer: Given the equation -16t+130t+1 your flare will be 200 feet or higher for 4.03 seconds.

This is an example for converting the standard form of a quadratic equation to vertex form:

Begin with an equation. We will use x^2 – x – 4.
We will begin by finding our vertex (h,k). To do this we must solve for x in the above equation where x = -b/2a.
1) x = -(-1)/2(1)
= -1
2) We will take the value x=3/2 and plug it back into our original equation to solve for y.
y = (-1)2 - (-1) - 4
= (-2) - (-1) - 4
= -5
3) We now know our vertex (-1, -5) and can know put our equation into vertex form.

In real life scenarios quadratics can be used to predict events that will happen in the future. When a ball is thrown in the air, quadratics can be used to find how high the ball went in the air, and the path the ball takes can be represented with a a graph of a parabola. Quadratics can also be used to predict the best amount of time to study for a student. For example, if there is a point where a student studies too much and their grade decreases if they study more, the student can find the best amount of time to study to get the best grade. They can also be used to predict temperatures, the predictions may not always be accurate, but with data from many years a quadratic equation can be a good indication of what the temperature will be on a given day.

In popular culture: Angry Birds is a popular game in which birds are flung at foes by the use of parabolas.
The game can only be mastered when one can master the arc of the parabola.

Angry Birds credited to Rovio Mobile. Picture from Symbian Tweet.

We see that the equation for the ball's parabola is the quadratic function y(t) = -0.704t^2 + 1.063t + 1.338. The general form of this equation is y(t)=-1/2gt^2 + v0t + h0 where g is the gravitational acceleration (32 ft/sec² or 9.8 m/sec²), t represents time, v sub 0 is our initial velocity and h sub 0 is our initial height before any time has elapsed.

Insert the quadratic function y(t) = -0.704t^2 + 1.063t + 1.338 into www.wolframalpha.com. Click show steps to see the process for solving the equation. The results are seen below. A graph is also provided where the landing and starting point of the ball from the ground can be seen. Using WolframAlpha provides students with an interactive way of learning. Using this program the students can understand how to solve the equation and visualize the results. You can also view the results at this link: http://www.wolframalpha.com/input/?i=y%3D-0.704t^2+%2B+1.063t+%2B+1.338.

There is also a quadratic equation calculator at the site: http://www.cool.math.com/students/calculators/source/quadratic.htm where students can insert the value of A, B, and C and the program will provide a solution for the two values of x. This is a useful tool for students to check their work.

Another web-site to visit is: http://www.purplemath.com/modules/quadform.htm where the quadratic formula is explored through various examples involving graphs and algebra. This is a neat thing for students to look at and possibly comment about for their homework. It explores the quadratic formula further and gives step-by-step explanation of what it does and how to do it.

This second section includes the use of a graph and algebra to find the x-intercepts on the graph. The website explains how to find these intercepts by using the graph and then explains the connection between graphing and solving an equation.

As you can see, the x-intercepts match the solutions, crossing the x-axis at x = –4and x = 1. This shows the connection between graphing and solving: When you are solving "(quadratic) = 0", you are finding the x-intercepts of the graph. This can be useful if you have a graphing calculator, because you can use the Quadratic Formula (when necessary) to solve a quadratic, and then use your graphing calculator to make sure that the displayed x-intercepts have the same decimal values as do the solutions provided by the Quadratic Formula. Note, however, that the calculator's display of the graph will probably have some pixel-related round-off error, so you'd be checking to see that the computed and graphed values were reasonably close; don't expect an exact match."

This next section provides the students with an equation. The website walks the student step-by-step through the quadratic formula and explains the implications of the results. Afterward, a graph is also provided, furthering the students' understanding of the solution.

More External links:
Here are a few sites that have some lessons involving quadratic formulas and another site that has a nice tool for making quadratic formula worksheets. The first site is a lesson plan directed towards middle school students but I think it is a good introduction/reminder for those in high school as well. The second site is Quadratic "Chutes and Ladders" and is a good game to help your students practice different ways of solving quadratic equations. The third site is a quadratic formula worksheet builder that allows you to input an amount and type of quadratic equations and it instantly produces a worksheet that can be printed off (not that it can't be done fairly easily on your own, but it's much quicker and provides an answer key). Enjoy!

F12: Ryan Garman, Michael Bartus, Brad Mesclier

W11: Colleen Condra, Amy Pickell, Josh Swackhammer, Abby Quisenberry

F11: Matt York, Joe Freiman

Quadratic Functions are very applicable to every day life. The best example of a quadratic function is the path of a ball being thrown into the air. Below is a diagram demonstrating this motion.

The arc as a whole can be plotted on a coordinate plane in the form of a quadratic graph. From this graph we can determine the equation of the parabola and the height of the ball. Using this equation we can determine the velocity and acceleration of the ball.

We will walk step-by-step through this process later in this section.

Sections:How the Quadratic Formula is related to a Quadratic Equation

The history of quadratics goes way back to 2000 BC when the Babylonians solved quadratics in radicals. The idea of quadratics has greatly developed since then. Chinese and Babylonian mathematicians solved quadratic equations by completing the square. When they solved quadratics they would focus on finding a length, and the length that was found was the root of the quadratic equation. The first steps towards finding a general formula to solve quadratic equations trace back around 300 BC, mathematicians Pythagoras and Euclid found a method to solve the quadratic equation by focusing strictly on geometry. Pythagoras noted that the ratios found a general procedure to solve the quadratic equation. During Euclid's time, there didn't exist the same notation for numbers and formulas that exist today, and because of that, it was not possible to compute the square root of any number by hand. In 800 AD, al-Khwarizmi, a Persian mathematician, created six new equations involving roots, squares of roots, and numbers which he used to solve quadratic equations, and derived the quadratic equation. The algebra used by him was entirely rhetorical, and he rejected negative solutions. These discoveries led Abraham bar Hiyya Ha-Nasi to publish a book with the complete solution of the quadratic formula. In the year 1545, Gerolamo Cardano, a mathematician from Italy and also one of the best algebraists of his time, collected all the works related to equations. Later in the year 1594, the quadratic formula was fist obtained by Simon Stevin, who was born in Bruges (now known as Belgium). Everything we know today stems off of these earlier discoveries. The quadratic formula as know it today isHistory:x1,2=(-b/2a) ± (1/2a)(b2-4ac)1/2,and this gives the solution to a generic quadratic equation of the form:

ax^2 +bx+c= 0.For more information about the history of quadratics and mathematics, go tohttp://mathworld.wolfram.com/QuadraticEquation.html, by Eric Weisstein

Standards and Principles:## The Common Core State Standards that apply:

In order for mathematics education to become more focused and coherent and to improve on overall mathematics achievement the states are implementing these common core state standards. The standards below are what Michigan requires of the mathematics curriculum:Include equations from linear and quadratic functions, and simple rational and exponential functions.xinto an equation of the form (x–p)2 =qthat has the same solutions. Derive the quadratic formula from this form.x2 = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them asa±bifor real numbersaandb.## The NCTM standards that apply:

## The HSCE standards that apply:

Back To Top

Different Forms of Quadratic Functions:(GeoGebra file available at GeoGebra Tube) A quadratic function has a "U" shaped graph called a parabola that can either open up or down. When the parabola opens upward the coefficient in front of the quadratic term is positive whereas when the parabola opens downward the coefficient is negative. A quadratic function has a standard form of

f(x)=ax^2 +bx + cwhereacannot equal 0. As seen in the above Geogebra applet when you manipulate the values ofa, b, andcyour parabola seems to change. Changing the value ofacauses your parabola to "shrink" or "stretch" in fact your parabola is actually going to infinity faster or slower than it was. Changingbcauses a change in the downward slope of your parabola. Changingcchanges where your parabola intersects the y-axis or it might also be called the height.(Geogebra file hosted at Geogebra Tube.) The vertex form a parabola is

f(x)=a(x - h)^2 +kwherehandkare the (x,y)coordinates of the vertex of the parabola the sign ofadetermines which way the parabola will open. Observe this by changing the values ofh,k, andain the Geogebra applet above. When the quadratic equation is set equal to 0 and we solve forxwe obtain the roots or zeros of the equation. These roots or zeros are when the values ofxmakef(x)=0 this is when our projectile is at a height of zero and crosses thex-axis.Back To Top

How the Quadratic formula is related to a Quadratic EquationIn this section, we will show how closely connected the quadratic formula to a quadratic equation step-by-step with the picture below as a reference so that you may see what is being done.

+ For the first step, we will start with an arbitrary equation, where

ais nonzero anda,b,andc, .which is stated in step one. Observe that our equation is a second degree polynomial equation. This means that we will have two solutions for our input (x). We will now subtractcfrom each side of the equation.+ For step two, we will divide both sides of the equation by

a.+ From step three, we want to complete the square. To do that, we must add

b2/4a2 to both sides of the equation.+ From step four, we will complete the square for the left side of the equation.

+ From step five, we will combine the like-terms on the right side of the equation.

+ From step six, we will take square roots of both sides of the equation.

+ From step seven, we simplify the equation.

+ From step eight, we will subtract both sides of the equation by

-b/2a+ From step nine, we will combine like-terms.

+ From step ten, there is no need to continue on because we have found the quadratic formula!!!!!

Now that you've seen how the equation works, let's see some examples and other ways quadratic equations are useful!

Photo Credit to Brad MesclierStandard and Vertex Form of Quadratic EquationsThe formula fora quadratic function is different from that of a linear function because it contains and x^2 term. Examples of quadratic functions include

f(x)= 2

(x^2)-4x-11g(x)=4-

(x^2)h(x)=(1/3)

(x^2)+(2/3)x+5The general form of a quadratic function, also known as the

standard form, is then as followsLet

a,b, andcbe constants with a /=/ 0. A function represented byf(x)=a(x^2)+bx+cis a

quadratic function.The domain of a quadratic function includes all real numbers and the leading coefficient of a quadratic function is

aas listed above. For the equations listed above the value ofais as followsa = 2a = -1a = (1/3)When a quadratic function f is expressed as f(x)=a(x^2)+bx+c, the coordinates of the vertex are not apparent. However, if

fis written asf(x)= a(x-h)^2 +k, then the vertex is located at(h,k). To justify that eh vertex is indeed(h,k), we can use an inequality of a. Ifa>0in the formf(x)= a(x-h)^2 +k, then the terma(x-h)^2is never negative and the minimum value off(x)isk. This value occurs whenx=hbecausef(h)=a(h-h)^2 +k = 0+k = kThus the lowest point on the graph of

f(x)=a(x-h)^2 +kwitha>0is(h,k) and because this graph is a parabola that opens upward, the vertex must be(h,k). You can prove this for a<0 in a similar fashion. So in summary the equation for the vertex form of a quadratic formula is as followsf(x)=a(x-h)^2+k

The x-intercepts(roots/zeros) can be found using many different methods. One requires us to our quadratic function you then set the factors equal to zero and solve for your variable. Another has you use a formula also known as the quadratic formula where you can plug in the values for a, b, and c, given by the quadratic function in standard form (ax^2+bx+c). This formula isExamples:Also, graphing and looking for the x-intercepts and completing the square are useful methods to keep in mind.

This is an example for solving a quadratic equation for the x-intercepts(roots/zeros):Begin with an equation. In this example we will use y = x^2 +4x+4.

For a quadratic it is important to factor the polynomial using the following steps.

1. Put y = 0 and insert this into the left side of the equation, such that

0 = x^2 +4x+4.

2. Determine how this can be factored. In this case we have to find two numbers that will equal 4 when multiplied, and also can be added together to get 4.

Options: (4 and 1), (2 and 2).

3. Once in factor form, we can set each factor equal to zero and solve for x, in this case

(x+2)(x+2) = 0

(x+2) = 0

(x+2) = 0

x = -2, -2

Therefore the x-intercepts, roots, or zeros are -2 and -2 in this case.

This is an example of solving using the quadratic formula:We begin with the equation x^2 +4x + 4. For the sake of the example and simplicity, we will pretend it can't be factored. We see the quadratic forumula:

x=

We have 3 variables:

a,b,andc. Our equation is in standard form: ax^2+bx+c = 0. In our equation, we can see that a=1, b= 4, and c= 4.Now we have everything we need; from here on out, it's just a matter of plugging in and making calculations.

Start by entering in values:

(-4±√(4^2 - (4*1*4) ))/(2*1)

Our determinant (the square root portion of the forumula) = 0.

This gives us 2 roots. -2, and -2 (since -4±0 = -4, and -4/2 = -2)

This checks with how we solved above with factoring. We could also check by looking at a graph of our function and making sure it crosses the x-axis only at -2.

Extension question: Our only root is -2, how is this possible with a parabola? Based on our coefficients, is it pointing up or down? Does this mean that the point (0,-2) is a maximum point or minimum?

Since there is only one point crossing the x-axis, this point must be a the vertex. Looking at our coefficient of 1, we can conclude that our parabola opens up at the point (0,-2). This is a global minimum.This is an example of solving a story problem using the quadratic formula:You were out hiking behind you house in the hills of North Dakota when you lost your footing and fell down a steep hill. When you reached the bottom you realized you broke your leg and there is no way you can make it back home by yourself. Being an experienced hiker you brought your flare gun in case of this occurrence. You know that the flare must be above 200 feet in the air of someone back home to see it. Given the equation -16

t+130t+1, given thattis in seconds, how long will the flare be in the air at 200 feet or higher?Start by setting your equation equal to 200 feet:

200=-16

t+130t+1Subtract 200 feet from both sides:

0=-16

t+130t-199Now use the quadratic formula to solve for

t:(-130±√(130^2 - (4*16*-199) ))/(2*-16)

Simplify:

(-130±√(4164))/(-32)

Now solve:

(-130+64.53)/-32=2.05

(-130-64.53)/-32=6.08

Therefore your flare will be at 200 feet or higher from 2.05 seconds to 6.08 seconds. Now find the total time the flare will be 200 feet or higher.

6.08-2.05=4.03

Answer: Given the equation -16

t+130t+1 your flare will be 200 feet or higher for 4.03 seconds.This is an example for converting the standard form of a quadratic equation to vertex form:Begin with an equation. We will use x^2 – x – 4.

We will begin by finding our vertex (h,k). To do this we must solve for x in the above equation where x = -b/2a.

1) x = -(-1)/2(1)

= -1

2) We will take the value x=3/2 and plug it back into our original equation to solve for y.

y = (-1)2 - (-1) - 4

= (-2) - (-1) - 4

= -5

3) We now know our vertex (-1, -5) and can know put our equation into vertex form.

f(x) = a(x - h)^2 + k

= 1(x-1)^2 - 5

A good example of solving a quadratic by completing the square can be seen here:

http://www.purplemath.com/modules/sqrquad.htm

Back To Top

Real Life ApplicationsIn real life scenariosquadratics can be used topredictevents that will happen in the future. When a ball is thrown in the air, quadratics can be used to find how high the ball went in the air, and the path the ball takes can be represented with a a graph of a parabola. Quadratics can also be used to predict the best amount of time to study for a student. For example, if there is a point where a student studies too much and their grade decreases if they study more, the student can find the best amount of time to study to get the best grade. They can also be used to predict temperatures, the predictions may not always be accurate, but with data from many years a quadratic equation can be a good indication of what the temperature will be on a given day.In popular culture: Angry Birds is a popular game in which birds are flung at foes by the use of parabolas.The game can only be mastered when one can master the arc of the parabola.

Angry Birds credited to Rovio Mobile. Picture from Symbian Tweet.Here is a fun video that connects the relationship of angry birds and quadratic functions:

http://www.youtube.com/watch?v=bsYLPlXl7VQ

A Few Useful Ideas

Let's examine this projection on a coordinate plane:Wolframalpha:http://scienceblogs.com/dotphysics/wp-content/uploads/2009/07/dotphys_2rwsw_2.jpg

We see that the equation for the ball's parabola is the quadratic function y(t) = -0.704t^2 + 1.063t + 1.338. The general form of this equation is y(t)=-1/2gt^2 + v0t + h0 where g is the gravitational acceleration (32 ft/sec² or 9.8 m/sec²), t represents time, v sub 0 is our initial velocity and h sub 0 is our initial height before any time has elapsed.

Insert the quadratic function y(t) = -0.704t^2 + 1.063t + 1.338 into www.wolframalpha.com. Click show steps to see the process for solving the equation. The results are seen below. A graph is also provided where the landing and starting point of the ball from the ground can be seen. Using WolframAlpha provides students with an interactive way of learning. Using this program the students can understand how to solve the equation and visualize the results. You can also view the results at this link: http://www.wolframalpha.com/input/?i=y%3D-0.704t^2+%2B+1.063t+%2B+1.338.

There is also a quadratic equation calculator at the site: http://www.cool.math.com/students/calculators/source/quadratic.htm where students can insert the value of A, B, and C and the program will provide a solution for the two values of x. This is a useful tool for students to check their work.

@davidwees davidweesTwitter:101 ways to use a quadratic equation in real life. Part 1 http://bit.ly/e0KHmm Part 2 http://bit.ly/haf5qZ #edchat #mathchat

Another web-site to visit is: http://www.purplemath.com/modules/quadform.htm where the quadratic formula is explored through various examples involving graphs and algebra. This is a neat thing for students to look at and possibly comment about for their homework. It explores the quadratic formula further and gives step-by-step explanation of what it does and how to do it.Purplemath:This second section includes the use of a graph and algebra to find the x-intercepts on the graph. The website explains how to find these intercepts by using the graph and then explains the connection between graphing and solving an equation.

As you can see, the

x-intercepts match the solutions, crossing thex-axis atx= –4andx= 1. This shows the connection between graphing and solving: When you are solving "(quadratic) = 0", you are finding thex-intercepts of the graph. This can be useful if you have a graphing calculator, because you can use the Quadratic Formula (when necessary) to solve a quadratic, and then use your graphing calculator to make sure that the displayedx-intercepts have the same decimal values as do the solutions provided by the Quadratic Formula.Note, however, that the calculator's display of the graph will probably have some pixel-related round-off error, so you'd be checking to see that the computed and graphed values were reasonably close; don't expect an exact match."

This next section provides the students with an equation. The website walks the student step-by-step through the quadratic formula and explains the implications of the results. Afterward, a graph is also provided, furthering the students' understanding of the solution.

More External links:Here are a few sites that have some lessons involving quadratic formulas and another site that has a nice tool for making quadratic formula worksheets. The first site is a lesson plan directed towards middle school students but I think it is a good introduction/reminder for those in high school as well. The second site is Quadratic "Chutes and Ladders" and is a good game to help your students practice different ways of solving quadratic equations. The third site is a quadratic formula worksheet builder that allows you to input an amount and type of quadratic equations and it instantly produces a worksheet that can be printed off (not that it can't be done fairly easily on your own, but it's much quicker and provides an answer key). Enjoy!

http://teachingtoday.glencoe.com/lessonplans/introducing-the-properties-of-quadratic-equations

http://www.learnnc.org/lp/pages/2981

http://www.theteacherscorner.net/printable-worksheets/make-your-own/math-worksheets/algebra/quadratic-equations.php

http://www.youtube.com/watch?v=43JvqjHtFXQFun youtube stuff:

http://www.yourteacher.com/search-keyword-results.phpFor some lesson plans for quadratic formulas:http://www.youtube.com/watch?v=tjroyVI8El4

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