# Safety_kit_2

```Max/min
Finding Roots
You should know the following about quadratic
functions:
 How to graph them
 How to find the vertex
 How to find the x- and y- intercepts
 How to find the equation from the pattern
 How to find the equation from the graph
 How to change from one form to another
There are basically two types of quadratic word problems:
you to find the
vertex
you to find the
roots
By giving you
the equation
There are basically two types of quadratic word problems:
you to find the
vertex
you to find the
roots
By
giving
you the
information to
find the equation
These are
the harder
ones!!
We have seen that the vertex of a quadratic in
general form is given by
b b

2a, f2a





The y-value of the vertex is either the
maximum value the function can have or it’s
the minimum value the function can have.
y-value’s a min when a >0
y-value’s a max when a<0
Example 1
A small business’ profits over the last year have been related to
the price of the only product. The relationship is
R(p) = -0.4p2 +64p-2400, where R is the revenue measured in
thousands of dollars and p is the price of the product
measured in dollars.
a. What price would maximize the revenue?
The word maximize screams
“FIND THE VERTEX!!”
Thus, a price of \$80
would maximize the
revenue.
b
2a
 64
p 
2 (  0 .4 )
64
p 
 0 .8
p  80
p 
Example 1
A small business’ profits over the last year have been related to
the price of the only product. The relationship is
R(p) = -0.4p2 +64p-2400, where R is the revenue measured in
thousands of dollars and p is the price of the product
measured in dollars.
b. What is the maximum revenue possible?
The answer to this question is
the y-value of the vertex.
The maximum revenue
is \$160 000.
p80
R
(80
)0.4(80
)264
(80
)240
R
(80
)0.4(6400
)5120
240
R
(80
)2560
5120
2400
R
(80
)160
Example 1
A small business’ profits over the last year have been related to
the price of the only product. The relationship is
R(p) = -0.4p2 +64p-2400, where R is the revenue measured in
thousands of dollars and p is the price of the product
measured in dollars.
c. How much money would they lose if they
gave the product away?
This question is talking about
a price of 0 or p = 0
lose \$2 400 000.
p

0
2
R
(
0
)

0
.4
(
0
)

64
(
0
)
24
R
(
0
)

2400
Example 2: no equation
A farmer needs to fence off his animals. He bought 800 m of
fencing and would like to maximize the area for his livestock.
According to regulations the pens need to find these relative
dimensions:
Kids
Pig Pen
Sheep
pen
What are the
dimensions of the
largest area?
Example 2: no equation
A farmer needs to fence off his animals. He bought 800 m of
fencing and would like to maximize the area for his livestock.
According to regulations the pens need to find these relative
dimensions:
Again the word
Maximum means
VERTEX. BUT…
Example 2: no equation
A farmer needs to fence off his animals. He bought 800 m of
fencing and would like to maximize the area for his livestock.
According to regulations the pens need to find these relative
dimensions:
We know:
The total fencing is
800m. This means
3w+3w+3w+8L+8L+w+2L=800
This equation has
two unknowns!!
or 10w + 18L =800
Example 2: no equation
A farmer needs to fence off his animals. He bought 800 m of
fencing and would like to maximize the area for his livestock.
According to regulations the pens need to find these relative
dimensions:
The question is
maximum area. So
let’s get an area
equation:
Area of a rectangle
A = (3w)(8L)
A = 24wL
The two equations are
A = 24wL
10w + 18L =800
To solve this system of equations,
use substitution.
Solve the linear equation (the one
without the multiplication of variables)
for one of the unknowns.
10w + 18L =800
10w =800 -18L
w = 80 -1.8L
Substitute this into the area
equation.
A = 24wL
A = 24(80-1.8L)L
A = 24L(80-1.8L)
A = 1920L-43.2L2
A = 1920L-43.2L2
b
2a
 1920
L 
2 (  43 . 2 )
 1920
L 
 86 . 4
L  22 . 22
L 
Remember, we’re
looking for the
MAX AREA. We’ll
find that with the
vertex.
The BEST L value to use is 22.22m
This leads to a w value of
w = 80 -1.8L
w = 80-1.8(22.22)
w= 40.00m
Example 2: no equation
A farmer needs to fence off his animals. He bought 800 m of
fencing and would like to maximize the area for his livestock.
According to regulations the pens need to find these relative
dimensions:
The dimensions that would
be the best are: w = 40.00m
and L = 22.22m
The largest area
would be:
A = 24wL
A = 24(40.00)(22.22)
A = 21331.2m2
A lifeguard has 75m of rope to section off the supervised
area of the beach. What is the largest rectangular
swimming area possible?
2w + L = 75
A = wL
L= 75-2w
A=w(75-2w)
A = 75w-2w2
A = 75(18.75)-2(18.75)2
A =1406.25-703.125
A= 703.125
b
2a
 75
w 
2( 2)
w  18 . 75
w 
We know that we can find the roots of a quadratic
function by setting one side equal to zero and
Factoring (sometimes)
Completing the square
(too long)
root formula
This ALWAYS
works for a
general form and
is easy to do.
2

b
b

4
ac
x
2
a
We know that we can find the roots of a quadratic
function by setting one side equal to zero and
Factoring (sometimes)
Completing the square
(too long)
root formula
This ALWAYS
works for a
general form and
is easy to do.
2

b
b

4
ac
x
2
a
Example 1
A duck dives under water and its path is described by
the quadratic function y = 2x2 -4x, where y represents
the position of the duck in metres and x represents
the time in seconds.
a. How long was the duck underwater?
The duck is no longer underwater when the
depth is 0. We can plug in y= 0 and solve
for x.
0  2x2  4x
0  2x(x  4)
0  2x
0  x4
So x = 0 or 4
The duck
was
underwater
for 4
seconds
Example 1
A duck dives under water and its path is described by
the quadratic function y = 2x2 -4x, where y represents
the position of the duck in metres from the water and
x represents the time in seconds.
52x24x
b. When was the duck at a depth of 5m?
We can plug in y= -5 and solve for x.
We cannot solve
this because
there’s a negative
number under the
square root.
We conclude that
the duck is never
5m below the
water.
02x24x5
 b  b 2  4ac
x
2a
 (4)  (4) 2  4(2)(5)
x
2(2)
4  16  40
4
4   24
x
4
x
Example 1
A duck dives under water and its path is described by
the quadratic function y = 2x2 -4x, where y represents
the position of the duck in metres from the water and
x represents the time in seconds.
We conclude that
b. When was the duck at a depth of 5m?
the duck is never
5m below the
We can check this by finding the
water.
minimum value of y.
 b
2 a
 ( 4 )
x 
2 (2 )
4
x 
4
x  1
x 
y 
2 (1 )
y 
 2
2
 4 (1 )
Example 1
A duck dives under water and its path is described by
the quadratic function y = 2x2 -4x, where y represents
the position of the duck in metres and x represents
the time in seconds.
c. How long was the duck at least 0.5m
below the water’s surface?
We can plug in y= -0.5 and solve for x.
This will give us the
times when the duck is
at 0.5 m below.
The duck was 0.5m
below at t = 0.14s
and at t = 1.87s
Therefore it was
below 0.5m for 1.73s

0
.52
x2
4
x
02
x2
4
x
0
.5
 b  b 2  4 ac
x
2a
 (  4 )  (  4 ) 2  4 ( 2 )( 0 . 5 )
x
2(2)
4  16  4
4
4  12
x
4
4  3 . 46
x
4
x  0 . 14 or 1 . 87 s
x
Example 2: no equation
A rectangular lawn measures 8m by 6m. The
homeowner mows a strip of uniform width around the
lawn, as shown. If 40% of the lawn remains unmowed,
what is the width of the strip?
40%
-6-2x-
The area of the
lawn is 8 x 6 =48
-----8-2x-----
40% of this is
unmowed:
48 x 0.40 = 19.2
The dimensions of
this unmowed
rectangle are
Example 2: no equation
A rectangular lawn measures 8m by 6m. The
homeowner mows a strip of uniform width around the
lawn, as shown. If 40% of the lawn remains unmowed,
what is the width of the strip?
40%
-6-2x-
So 19.2 = (8-2x)(6-2x)
-----8-2x-----
We need to solve for x
Make one side equal to 0 and
use the quadratic root formula
But first we FOIL it out
Example 2: no equation
A rectangular lawn measures 8m by 6m. The
homeowner mows a strip of uniform width around the
lawn, as shown. If 40% of the lawn remains unmowed,
what is the width of the strip?
40%
-6-2x-
So 19.2 = (8-2x)(6-2x)
-----8-2x-----
19
.248
16
x12
x4x2
19
.24x2 28
x48
04x2 28
x48
19
.2
04x2 28
x28
.8
Example 2: no equation
A rectangular lawn measures 8m by 6m. The
homeowner mows a strip of uniform width around the
lawn, as shown. If 40% of the lawn remains unmowed,
what is the width of the strip?
40%
-6-2x-
0 = 4x2-28x+28.8
-----8-2x-----
The mowed strip has
a width of 1.25m
b b2 4ac
x
2a
(28)  (28)2 4(4)(28.8)
x
2(4)
2817.98
x
8
x 1.25m or 11.45m
Example 2: no equation
A rectangular lawn measures 8m by 6m. The
homeowner mows a strip of uniform width around the
lawn, as shown. If 40% of the lawn remains unmowed,
what is the width of the strip?
40%
-6-2x-
Let’s check:
-----8-2x-----
If x = 1.25m then the
length is
8 – 2x
= 8 -2(1.25)
=5.5m
width is
6 – 2x
= 6 -2(1.25)
=3.5m
A = (5.5)(3.5)
A =19.2
Which was 40% of
the total area!
Roots Word Problems
Example 3: no equation
Two numbers have a difference of 18. The sum of
their squares is 194. What are the numbers?
Let’s define our variables:
S = one of the numbers
G= the other number
The question indicates two
equations relating these two
variables.
S – G = 18
S2+ G2 = 194
Again, we have a substitution
S=18+G
situation. Solve the simpler
equation for a variable and plug it
in to the other equation.
(18+G)2 + G2 = 194
Roots Word Problems
Example 3: no equation
Two numbers have a difference of 18. The sum of
their squares is 194. What are the numbers?
(18+G)2 + G2 = 194
We need to solve this equation for G.
Use the quadratic root formula
Let’s FOIL and make one
side equal to 0.
324+36G+
G2
+
G2
= 194
2G2 +36G+324-194 =0
2G2 +36G+130=0
b  b2  4ac
G
2a
(36)  (36)2  4(2)(130)
G
2(2)
3616
G
4
G  5 or 13
S = 18 + (-5) = 13 or S = 18+(-13)=5
Roots Word Problems
Example 3: no equation
Two numbers have a difference of 18. The sum of
their squares is 194. What are the numbers?
(18+G)2 + G2 = 194
The numbers are either -5 and 13
or -13 and 5
Roots Word Problems: Try one
A rectangle is 8 feet long and 6 feet wide. If the same
number of feet increases each dimension, the area of
the new rectangle formed is 32 square feet more than
the area of the original rectangle. How many feet
increased each dimension?
So
80 = (6+x)(8+x)
80=48+14x+x2
0=x2 +14x-32
x =2 or x = -16
The new are is
32+6x8
=32+48
=80
The dimensions
of the new
rectangle are
6+x and 8+x
A ball is thrown and follows the path described by the function
h(t) = -5t2 +20t +1, where h is the height of the ball and t is the
time since the ball was released.
0 = -5t2 +20t -2.5
a) When was the ball at a height of 3.5m?
This question is looking for ‘t’ so it
gives a specific h. In this case h = 3.5.
20 202 4(5)(2.5)
t
2(5)
20 350
10
2018.71
t
10
t
3.5 = -5t2 +20t +1
We will solve this by setting one side
equal to zero and using the
One time is on the way up and
the other is on the way down.
t
2018.71
10
or
t  0.129s or t  3.871s
t
2018.71
10
A ball is thrown and follows the path described by the function
h(t) = -5t2 +20t +1, where h is the height of the ball and t is the
time since the ball was released.
0 = -5t2 +20t -2.5
a) When was the ball at a height of 3.5m?
The ball is at a height of 3.5m at two
times: at t= 0.129s and at t = 3.871s
20 202 4(5)(2.5)
t
2(5)
20 350
10
2018.71
t
10
t
t
2018.71
10
or
t  0.129s or t  3.871s
t
2018.71
10
A ball is thrown and follows the path described by the function
h(t) = -5t2 +20t +1, where h is the height of the ball and t is the
time since the ball was released.
b) How high is the ball after 4.0s?
This question is looking for h
given a value of t.
t = 4.0
h = -5(4)2 +20(4) +1
h = -80 +80 +1
h=1
After 4.0 seconds in
the air, the ball is 1 m
off the ground.
A ball is thrown and follows the path described by the function
h(t) = -5t2 +20t +1, where h is the height of the ball and t is the
time since the ball was released.
c) What is the ball’s maximum height?
The question is asking for height so I
must know the time. Do I?
The word MAXIMUM
screams VERTEX!!
I do know the time
value…
It’s
b
t
2a
The ball
reaches its
maximum
height 2.0
seconds after
being thrown
h = -5t2 +20t +1
 b
2a
 ( 20 )
t 
2 ( 5)
 20
t 
 10
t  2
t 
The max height:
h
5
(2
)220
(2
)1
h
20
40
1
h21
The maximum
height is 21 m
A ball is thrown and follows the path described by the function
h(t) = -5t2 +20t +1, where h is the height of the ball and t is the
time since the ball was released.
0 = -5t2 +20t +1
d) When does the ball hit the ground?
This question is asking for the time so
I must know the height.
20 202 4(5)(1)
t
2(5)
The height is 0 – hitting the ground!
t
Time can’t be negative
so this cannot be an
20 420
10
2020.49
t
10
t
2020.49
10
or
t
t  0.049s or t  4.049s
2020.49
10
A ball is thrown and follows the path described by the function
h(t) = -5t2 +20t +1, where h is the height of the ball and t is the
time since the ball was released.
e) From what height was the ball thrown?
This question is asking for the height
so I must know the time.
The time is 0 – just before it is thrown!
h = -5(0)2 +20(0)+1
h =1
```