Height of a Zero Gravity Parabolic FlightMath 1010 Intermediate Algebra Group ProjectLillian Fritzel

#1 Identify the 3 by 3 systems of equations for a, b and c

#2Solve the system

#3Use is to form the quadratic

formula

#4Find the

maximum value of the quadratic function

#5Graph

Have you ever wondered what it might feel like to float is space? In this presentation I will demonstrate how to find a quadratic formula from data and how to be able to Understand how long you may have to feel the sensation that you are weight less!

STEP#1Understanding the Data

Height of a Zero-G Flight t Seconds AfterStarting a Parabolic Flight Path

Time t on seconds

2 20 40

Height h in feet

23645 32015 33715

In this lab we took a look at a parabolic path and determined the maximum altitude the plane reaches during the parabolic flight.

To do this we first looked at our data given and indentified the x and y axis to then plug in our points to form three system equations to the model a quadratic model for the flight.

STEP#2 From the data we can now plug it into a formula

cbtath 2

cba )2()2(23645 2

cba )20()20(32015 2 c b a ) 40 ( ) 40 ( 337152

We switched the equation around so that the variables are on the left side of the equality * Notice the location of (2), (20), (40) as well as 23645, 32015, and 33715*

We now have a 3 by 3 system to solve (Step#2 continued)

2231568510 2 xxy

4a+2b+c=23645 (1)

400a+20b+c=32015 (2)

1600a+40b+c=33715 (3)

(-1)-4a-2b-c=-23645+4

00a+20b+c=32015= 396a+18b=8370

(4)

(-1)-400a-20b-c=-32015

+1600a+40b+c=33715

= 1200a+20b=1700(5)

The next step is to first eliminate the c variables from the equations

I multiply (-1) by line (1) and add line (2) in order to cancel c to create line (4)

Then I multiply (-1) by line (2) and add line (3) to cancel c to create line (5)

-1200(396a+18b=83

70)=

-475200a-21600b= -10044000

396(1200a+20b=1700)=

475200a+7920b=673200

b= -9370800/-13680b= 685

(6)

Cancel c solve for b(step#2 cont.)

Combine line 4 With line 5 ) We then canceled a to solve for b. To do this you can multiply each equation by it’s a variable and add together. (a) will then cancel and you are left with two equations equal to b. From here you would set both equations to b and divide to find that b=685

Substitute b to Solve for a then

substitute a to solve for c

(step#2)1200a+20(685)=17001200a+13700=1700

-13700=-12000/1200

a= -10

4(-10)+2(685)+c=23645-40+1370+c=23645 -1330

c= 22315

STEP#3 Write the quadratic functionFrom this point we plugged a, b, c

Into the quadratic formula to the right-

to get x intercepts for plotting the graph latter on

Our quadratic model/function is the following quadratic equation. Since there is a (-) in front of the mx+b [10] then it contains a maximum or the graph may look like an arched bridge

a=-10

b=685

c=22315

2231568510 2 xxy

Step#4 Find the Maximum value of the quadratic model To find the

Maximum we must use a formula that finds a vertex of a quadratic function.

We plug 34.25 into quadratic function

This also gives us the axis of symmetry

a

bx

2

25.34

)10(2

685

22315)25.34(685)25.34(10 2 v 625.34045

625.34045,25.34

STEP#5 Graph to show the parabolic flight

(34.25,34045.625) is the axis of symmetry as well as the Maximum

*special thanks to http://www.webgraphing.com/graphing_basic.jsp

625.34045,25.34

Review•STEP 2•Substitute b=685 into equation (5)

•Solve for a

a= -10

•STEP 1•Combining (4) + (5) after canceling for a

•Sove for b

b= 685 •STEP 3

•Substitute a=-10 and b=685 into oringinal equation (1)

•Solve for c

c=22315

*Find data point to create 3 by 3 system*Solve the system*Write a quadratic function*Find Maximum of function*Graph the parabolic flight

Thank you for looking at my presentation!

Created by:Lillian Fritzel

Math 1010