1.  Liela is standing on the opponents 40 yard line. She throws a pass
toward the goal line. The ball is 2 meters above the ground when
she lets go. It follows a parabolic path, reaching its highest point,
14 meters above the ground, as it crosses the 20 yard line.
a. Define your variables. You will receive NO CREDIT for a problem
if the variables are not defined!!!!
Let x = the number of yards from the goal line.
Let y = the no. of meters the ball is above the ground.
# means a pound sign - DO NOT USE IT to
abbreviate the word number, use no. .)
b.  Decide on the proper general equation either linear or
quadratic. (If quadratic decide between vertex form or
standard form).
It follows a parabolic path
It is quadratic.
reaching its highest point
Use vertex form.
1.  Liela is standing on the opponents 40 yard line. She throws a pass
toward the goal line. The ball is 2 meters above the ground when
she lets go. It follows a parabolic path, reaching its highest point,
14 meters above the ground, as it crosses the 20 yard line.
c. Write your coordinate points..
vertex: (20, 14)
other point: (40, 2)
d. Write the particular equation of the parabola.
2 - 14 = a(40 - 20)2
Unless you are told otherwise you leave
the equation in vertex form. Putting it in
standard graphing form, may be useful,
but it is not always necessary.
-12 = a(20)2
y - 14 = -.03(x2 - 40x + 400)
-12 = 400a
y - 14 = -.03x2 + 1.2x - 12
a = -0.03
y = -.03x2 + 1.2x + 2
y - k = -.03(x - h)2
The y-intercept is (0, 2). This could
have been found by plugging 0 in for x
and solving for y.
y - k = a(x - h)2
y - 14 = a(x - 20)2
y - 14 = -.03(x - 20)2
1.  Liela is standing on the opponents 40 yard line. She throws a pass
toward the goal line. The ball is 2 meters above the ground when
she lets go. It follows a parabolic path, reaching its highest point,
14 meters above the ground, as it crosses the 20 yard line.
y = -.03x2 + 1.2x + 2
e. How high is the ball when it crosses the 10 yard line?
y = -.03(10)2 + 1.2(10) + 2
y = 11
The ball is 11 meters high when it crosses the 10 yard line.
(You MUST LABEL ALL ANSWERS to word problems!!!!)
1.  Liela is standing on the opponents 40 yard line. She throws a pass
toward the goal line. The ball is 2 meters above the ground when
she lets go. It follows a parabolic path, reaching its highest point,
14 meters above the ground, as it crosses the 20 yard line.
y = -.03x2 + 1.2x + 2
f.
If nobody caches the ball, approximately where will it hit the
ground?
2
!b
±
b
! 4ac
0 = -.03x2 + 1.2x + 2
x=
2a
b2 - 4ac
!1.2 ± 1.68
1.22 - 4(-.03)(2)
x=
2(!.03)
1.44 + .24
1.68
The ball hits the ground
1.6 yards into the end
zone.
"1.2 ± 1.2961
x!
".06
x ! 41.60 or x ! "1.60
1.  Oren and Liela are trapped in a room at a space station. The room
is 20 meters long and 15 meters wide. But the length is decreasing
linearly with time at a rate of 2 meters per minute, and the width is
increasing linearly with time at a rate of 3 meters per minute.
a. Let L(t) and W(t) be the length and width of the room,
respectively, in meters. Let t be the number of minutes since
the room was 20 by 15. Write particular equations for L and W.
L(t) = -2t + 20
W(t) = 3t + 15
b. Let A(t) be the number of square meters of floor area in the
room. Write the particular equation for function A.
A(t) = (-2t + 20)(3t + 15)
= -6t2 + 30t + 300
1.  Oren and Liela are trapped in a room at a space station. The room
is 20 meters long and 15 meters wide. But the length is decreasing
linearly with time at a rate of 2 meters per minute, and the width is
increasing linearly with time at a rate of 3 meters per minute.
A(t) = -6t2 + 30t + 300
c.  Does the area of the room maximize or minimize for a positive
value of t If so, when and how large or small?
y - 300 = -6t2 + 30t
y - 300 = -6(t2 - 5t)
y - 300 + -6(6.25) = -6(t2 - 5t + 6.25)
y - 262.5 = -6(t - 2.5)2
At a time of 2.5 minutes the room will reach a maximum
area of 262.5 feet.
1.  Oren and Liela are trapped in a room at a space station. The room
is 20 meters long and 15 meters wide. But the length is decreasing
linearly with time at a rate of 2 meters per minute, and the width is
increasing linearly with time at a rate of 3 meters per minute.
A(t) = -6t2 + 30t + 300
c.  When will the area of the room be zero?
0 = -6t2 + 30t + 300
b2 - 4ac
302 - 4(-6)(300)
900 + 7200
8100
0 = t2 - 5t - 50
0 = (t - 10)(t + 5)
t = 10 or t = -5
The room will have an area of zero after 10 minutes.
In a problem that looks at the change in height over time (such as throwing
a rock, launching a rocket, jumping off a diving board, etc.) the equation of
y = ax2 + bx +c has special values.
b is the initial vertical velocity. If and object is thrown or launched upward, b
is positive. If it is thrown or launched downward, b is negative. If it is
simply dropped, b is zero.
The leading coefficient, a, is acceleration (usually a measure of the force of
gravity). The number is negative because gravity pulls downward. Gravity
is either –32 feet per seconds squared or -9.8 meters per seconds squared.
c is the initial height. An object always starts its path at time zero, which is
the y-intercept.
An equation in physics for parabolic curves acted on by gravity is:
h(t) = -(1/2)gt2 + v0t + h0.
Where h(t) is a function of height that depends on time; t is time; g is
gravity, v0 is the initial velocity, and h0 is the initial height.
Use the information above to solve the following questions.
1.  Liela is standing on the roof of Newton South High School 30 feet
above the ground. She uses a sling shot to launch a rock up into the
air with an initial velocity of 55 feet per second.
a) Write the quadratic function that models this situation.
Let t = time in seconds
Let h(t) = height in feet.
h(t) = -(1/2)gt2 + v0t + h0
g = 32 ft/s2
v0 = 55 ft/s
h0 = 30
(Unless you are told the height of the person, you
can ignore it.)
h(t) = -(1/2)(32)t2 + 55t + 30
h(t) = -16t2 + 55t + 30
Use the information above to solve the following questions.
1.  Liela is standing on the roof of Newton South High School 30 feet
above the ground. She uses a sling shot to launch a rock up into the
air with an initial velocity of 55 feet per second.
h(t) = -16t2 + 55t + 30
b) Will the rock ever reach a height of 90 feet?
h(t) = 90
90 = -16t2 + 55t + 30
0 = -16t2 + 55t - 60
b2 - 4ac = 552 - 4(-16)(-60) = -815
Since there is no real solution, the rock will never reach this height.
Use the information above to solve the following questions.
1.  Liela is standing on the roof of Newton South High School 30 feet
above the ground. She uses a sling shot to launch a rock up into the
air with an initial velocity of 55 feet per second.
h(t) = -16t2 + 55t + 30
c) What is the maximum height of the rock?
y - k = a(x - h)2
h = -b/2a = -55/2(-16) = 55/32 ≈ 1.71875 ≈ 1.7
k = h(55/32) = -16 (55/32) 2 + 55 (55/32) + 30 = 4945/64 ≈ 77.2656 ≈ 77.3
The maximum height is about 77.3 feet.
Use the information above to solve the following questions.
1.  Liela is standing on the roof of Newton South High School 30 feet
above the ground. She uses a sling shot to launch a rock up into the
air with an initial velocity of 55 feet per second.
h(t) = -16t2 + 55t + 30
d) When will the rock hit the ground?
h(t) = 0
0 = -16t2 + 55t + 30
b2 - 4ac = 552 - 4(-16)(30) = 4945
!b ± b 2 ! 4ac !55 ± 4945 !55 ± 70.3207
t=
=
"
2a
!32
!32
t " !0.4788 or t " 3.9163
t " 3.9
It will hit the ground after about 3.9 seconds.
Reminders:
1.  # is not an abbreviation for the word number. You will lose points for
using it.
2.  & is not an abbreviation for the word and. You will lose points for using
it.
3.  Mixed fractions are not allowed.
Use of them will result in no credit for the problem.
4. In order to receive credit on a graphing problem, the graph must:
a. Include a properly drawn scale.
b. Have its axis labeled properly.
c. Have all straight lines drawn with a ruler.
5.  All answers to word problems must include a label. Otherwise they are
incorrect and will graded as a wrong answer.
6.  You may approximate answers to word problems to whatever decimal
makes sense.
Using the graphing calculator
I will give you a packet on everything you need when I see you next.