Math 20-1 Chapter 3 Quadratic Functions
3.2 Quadratic Standard Form
Teacher Notes
3.2 Quadratic Functions in Standard Form
Express y  2( x  3) 2  5 in standard form.
y  2 x 2  12 x  23
Chapter 2
f(x) = ax + bx + c is called the ( standard form / vertex form )
of a quadratic function.
Parameter(s)
3
( a / b / c ) determines whether the graph opens
upward or downward.
Parameter(s) ( a / b / c ) influences the position of the
graph.
Parameter(s) ( a / b / c ) is the y-intercept of the graph.
3.2.1
Listing Characteristics of Quadratic Function using Technology
y  3x  18x  23
2
Vertex
(3, -4)
Axis of Symmetry
x=3
Max/Min
Min of y = -4
x - intercept(s)
(1.8, 0)
y- intercept
(0, 23)
Domain
x R
Range
y ≥ -4
(4.1, 0))
3.2.2
McGraw Hill Page 168 Example 2
A frog sitting on a rock jumps
into a pond. The height, h, in
cm, of the from above the
surface of the water as a
function of time, t, in sec, since
it jumped can be modelled by
the function h(t )  490t 2  150t  25
a) Graph the function.
b) What is the value of the y-intercept?
25cm
What does it represent?
At time, t = 0 the height of
the frog is 25cm.
The from jumped from a height of 25cm.
3.2.3
c) What characteristic of the graph
represents the the maximum height
reached by the frog? vertex
(0.2, 36.5)
d) What is the maximum height?
36.5 cm
e) When does the from reach it’s max
0.2 sec
height?
f) How long is the frog in the air?
0.4 sec
i) What is the height of the
frog at 0.25 s? 31.9 cm
j) At what time is the
g) What is the domain of the situation? height of the frog 30 cm ?
[0, 0.4]
0.038 and 0.27 sec
h) What is the range of the situation?
[0, 36.5]
3.2.4
Problem 2 Max Area of Rectangle
Find the dimensions of a rectangular lot of maximum
area that can be enclosed by 1000 m of fence.
500 - w
w
w
Fencing:
1000 = 2w + 2(length)
1000 – 2w = 2length
length = 500 - w
V(250, 62 500)
For the area:
A = w(500 -w )
A = 500w -w2
Therefore, the dimensions of the lot are
250 m by 250 m.
w = 250
3.2.5
Problem 3 Max Area of Rectangle with Missing Side
Find the dimensions of the lot if only three sides need to be
enclosed by the 1000 m of fence. Find the maximum area.
x
x
1000 - 2x
For the length of fence:
1000 = 2x + length
Length = 1000 - 2x
V(250, 125 000)
x = 250
For the area:
A = x(-2x + 1000)
A = -2x2 + 1000x
length = -2x + 1000
length = -2(250) +1000
length = 500
Therefore, the dimensions are
250 m x 500 m.
The maximum area is
125 000 m2.
3.2.6
Problem 4 Max Area of Rectangle with Divisions
A rectangular field is to be enclosed by a fence and then
divided into three smaller plots by two fences parallel to one
side of the field. If there are 900 m of fence to be used, find
the dimensions and the maximum area of the field.
length
x
x
x
x
For the fence:
900 = 4x + 2length
450 = 2x + length
length = -2x + 450
For the area:
A = xlength
A = x(-2x + 450)
A= -2x2 + 450x
-2x + 450
V 112.5, 25312.5
y = -2x + 450
y = -2(112.5) + 450
y = 225
Therefore the dimensions of the
Field are 112.5 m x 225 m.
The maximum area of the field is
25 312.5 m2.
3.2.7
Problem 5 Revenue
How many pear trees should be planted to maximize
the revenue from an orchard for one year?
Research for an orchard has shown that, if 100 pear
trees are planted, then the annual revenue is $90 per
tree. The annual revenue per tree is reduced by
$0.70 for every additional tree planted.
Revenue = ($) (#)
Revenue =  90 100 
Maximum Revenue =
 90  0.7 x 100  1x 
What does x represent?
Vertex (14.28, 9142.86)
14 trees should be planted to maximize the revenue
Page 174:
1, 3, 5a,b, 7, 12, 15, 17, 23
3.2.8
Interpret a Graph: Javelin
Dave threw a javelin from ground level. The javelin travelled on a
parabolic path that could be defined by the equation
h = -0.5x2 + 6x +1, where h is the height that the javelin reaches in
metres, and x is the horizontal distance the javelin travels in metres.
vertex (6, 19)
From ground level, what is the
height
maximum height the javelin
reached? Maximum height is 19 m.
To reach maximum height, what
Horizontal distance
horizontal distance must the javelin
travel? The javelin must travel 6 horizontal metres.
What horizontal distance did the javelin travel in total?
The javelin travelled 12 m horizontally.
From what height was the javelin thrown?
1m
3.2.5
Interpreting a Graph
A flare pistol is fired in the air. The height of the flare above the
ground is a function of the elapsed time since firing.
vertex at (10, 500)
2
h = -5t + 100t
What is the maximum
height reached by the
flare?
height
Maximum height is at 500 m.
How long is the flare in the air?
time
The flare is in the air for 20 seconds.
3.3A1.11