Do Now:
WRITE THE EQUATION
FOR THE QUADRATIC
FORMULA FROM
MEMORY!!!!
Take out HW-hand in 

CW: Quadratic Formula
Imaginary Numbers!
We will
 Investigate how to find
the equation of a
quadratic sketch
HW: QUADRATIC
Formula WORKSHEET –
odds only!!!
 Intro:
We already know the
standard form of a quadratic
equation is:
y = ax2 + bx + c
 The coefficients are: a , b, c
 The
variables are: y, x
 The
ROOTS (or
solutions) of a
polynomial are
its x-intercepts
 The
x-intercepts
occur where y =
0.
Roots
Example: Find the
roots: y = x2 + x - 6
 Solution: Factoring:

y = (x + 3)(x - 2)
 0 = (x + 3)(x - 2)


The roots are:
x = -3; x = 2
Roots

After centuries of
work,
mathematicians
realized that as long
as you know the
coefficients, you can
find the roots of the
quadratic. Even if it
doesn’t factor!
y  ax 2  bx  c, a  0
b  b 2  4ac
x
2a

Watch this:
http://www.youtube.com/watch?v=MUw9C1p5Y0g&f
eature=related
(another time…)
http://vihart.com/doodling/

Extra Credit if you create a video/song/anything
creative!! You have 2 weeks!
Solve: y = 5x  8x  3
a  5, b  8, c  3
2
8  64  60
x
10
2
b  b  4ac
8

4
x
x
2a
10
2
(8)  (8)  4(5)(3)
8 2
x
x
2(5)
10
8 2
x
10
8  2 10
x
 1
10
10
82 6 3
x
 
10 10 5
Roots



 




2
 
 
 
 
y  5 35  8 35  3
y  5 9 25  24 5  3
45
24  3
y


2
25
5
y  5(1)  8(1)  3
9
24
15
y



y  583
5
5
5
y0
y0
Plug in your
answers for x.
If you’re right,
you’ll get y = 0.
2
Solve : y  2x  7x  4
DO NOW ** WRITE THE
EQUATION FOR THE
QUADRATIC FORMULA FROM
MEMORY then solve the
quadratic above!!!!
Class Work:
AS PROMISED….
Imaginary numbers!!



WE will:
Solve Quadratics using
quadratic formula!!
Check out the world of
IMAGINARY NUMBERS!!!
HW: QUADRATIC Formula
WORKSHEET – 1-10!!!
7

49

32
Solve : y  2x  7x  4 x 
4
a  2, b  7, c  4
7  81
x
2
b  b  4ac
4
x
7  9 2 1
2a
x
x 
4
4 2
2
(7)  (7)  4(2)(4)
16
x
x
 4
2(2)
4
2
Remember: All the terms must be on one
side BEFORE you use the quadratic
formula.
•Example: Solve 3m2 - 8 = 10m
•Solution: 3m2 - 10m - 8 = 0
•a = 3, b = -10, c = -8
Free Fallin’
The acceleration of
gravity (g) for objects in
free fall at the earth's
surface is 9.8 m/s2.
Galileo found that all
things fall at the same
rate.
Free Fall
The rate of falling
increases by 9.8 m/s
every second.
Height = ½ gt2
For example:
½ (9.8 )12 = 4.9 m
½(9.8)22 = 19.6 m
½ (9.8)32 = 44.1 m
½ (9.8)42 = 78.4 m
Free Fall
A ball thrown horizontally
will fall at the same rate as
a ball dropped directly.
Free Fall
A ball thrown into the air
will slow down, stop, and
then begin to fall with the
acceleration due to gravity.
When it passes the
thrower, it will be traveling
at the same rate at which it
was thrown.
Free Fall
An object thrown upward at an angle to the ground follows a curved
path called a parabola.
Air Resistance
• In air…
– A stone falls faster than a
feather
• Air resistance affects
stone less
• In a vacuum
– A stone and a feather will fall
at the same speed.
Air Resistance
• Free Fall
– A person in free fall reaches a
terminal velocity of around 54
m/s
– With a parachute, terminal
velocity is only 6.3 m/s
• Allows a safe
landing



The initial velocity is the coefficient for the middle
term, and the initial height is the constant term.
The coefficient on the leading term comes from the
force of gravity. This coefficient is negative, since
gravity pulls downward, and the value will either be
"4.9" (if your units are "meters") or "16" (if your
units are "feet"). In general, the format is:
s(t) = –gt2 + v0t + h0

...where "g" here is the "4.9" or the "16" derived
from the value of the force of gravity (technically,
it's half of the force of gravity, but you probably
don't need to know that right now), "v0" ("veenaught", or "vee-sub-zero") is the initial velocity,
and "h0" ("aitch-naught", or "aitch-sub-zero") is the
initial height.


An object is launched at 19.6 meters per second
(m/s) from a 58.8-meter tall platform. The equation
for the object's height s at time t seconds after
launch is s (t) = –4.9t2 + 19.6t + 58.8, where s is in
meters. When does the object strike the ground?
What do we know vs. What do we need to know? (3
min)
I'll set s equal to zero, and solve:
 0 = –4.9t2 + 19.6t + 58.8
0 = t2 – 4t – 12
0 = (t – 6)(t + 2)
Then t = 6 or t = –2. The second solution is from two
seconds before launch, which doesn't make sense in this
context. (It makes sense on the graph, because the line
crosses the x-axis at –2, but negative time won't work in
this word problem.) So "t = –2" is an extraneous solution,
and I'll ignore it.
The object strikes the ground six seconds after launch.




An object in launched directly upward at 64 feet per
second (ft/s) from a platform 80 feet high. What
will be the object's maximum height? When will it
attain this height?
Think Pair Share (3 min)
What do I know?
What do I need to find out??

The initial height is 80 feet above ground and the
initial speed is 64 ft/s. Since my units are "feet",
then the number for gravity will be 16, and my
equation is:
s(t) = –16t2 + 64t + 80

Yes?



They want me to find the maximum height. For a
negative quadratic like this, the maximum will be at
the vertex of the upside-down parabola. So they
really want me to find the vertex. From our past
experience with graphing, I know how to find the
vertex;
in this case, the vertex is at (2, 144):
x = –b/2a = –(64)/2(–16) = –64/–32 = 2
Y = s(2) = –16(2)2 + 64(2) + 80 = –16(4) + 128 + 80
= 208 – 64 = 144
But what does this vertex tell me? According to my
equation, I'm plugging in time values and extracting
height values, so the input "2" must be the time and
the output "144" must be the height.
It takes two seconds to reach the maximum height of
144 feet.
2  4  84
x
6
2  88
x
6
2
2  4 • 22
b  b  4ac
x
x
6
2a
2
(2)  (2)  4(3)(7) x  2  2 22
x
6
1  22
2(3)
x
3
 Solve:
3x2 = 7 - 2x
 Solution: 3x2 + 2x - 7 =
0
 a = 3, b = 2, c = -7