Math 20-1 Chapter 3 Quadratic Functions

3.2 Quadratic Standard Form

Teacher Notes

3.2.1

3.2 Quadratic Functions in Standard Form

Chapter

3Parameter(s) ( 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.

f(x) = ax2 + bx + c is called the ( standard form / vertex form )

of a quadratic function.

22( 3) 5y x Express in standard form.22 12 23y x x

Listing Characteristics of Quadratic Function using Technology

Vertex

Axis of Symmetry

Max/Min

x - intercept(s)

y- intercept

Domain

Range

(3, -4)

x = 3

Min of y = -4

(1.8, 0) (4.1, 0))

(0, 23)

x R

.

y 3x2 18x 23

y ≥ -4 3.2.2

McGraw Hill Page 168 Example 2A 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

2( ) 490 150 25h t t t

b) What is the value of the y-intercept?a) Graph the function.

What does it represent?

25cm

At time, t = 0 the height ofthe 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

d) What is the maximum height?

(0.2, 36.5)

e) When does the from reach it’s max height? 0.2 sec

36.5 cm

f) How long is the frog in the air?0.4 sec

g) What is the domain of the situation?[0, 0.4]

h) What is the range of the situation?[0, 36.5]

i) What is the height of the frog at 0.25 s? 31.9 cm

j) At what time is the height of the frog 30 cm ?

0.038 and 0.27 sec

3.2.4

Find the dimensions of a rectangular lot of maximum area that can be enclosed by 1000 m of fence.

w

500 - w

Fencing: For the area:

A = w(500 -w )A = 500w -w2

1000 = 2w + 2(length)1000 – 2w = 2length length = 500 - w

V(250, 62 500) Therefore, the dimensions of the lot are250 m by 250 m.

Problem 2 Max Area of Rectangle

w = 250

3.2.5

w

Find the dimensions of the lot if only three sides need to beenclosed by the 1000 m of fence. Find the maximum area.

x x

1000 - 2x

For the length of fence: 1000 = 2x + length

For the area:A = x(-2x + 1000)A = -2x2 + 1000x

V(250, 125 000) Therefore, the dimensions are250 m x 500 m.

The maximum area is125 000 m2.

Problem 3 Max Area of Rectangle with Missing Side

length = -2x + 1000length = -2(250) +1000length = 500

x = 250

3.2.6

Length = 1000 - 2x

A rectangular field is to be enclosed by a fence and thendivided into three smaller plots by two fences parallel to oneside of the field. If there are 900 m of fence to be used, findthe dimensions and the maximum area of the field.

x x x x

-2x + 450

lengthFor the fence:900 = 4x + 2length450 = 2x + length length = -2x + 450

For the area:A = xlengthA = x(-2x + 450)A= -2x2 + 450x

Therefore the dimensions of theField are 112.5 m x 225 m.

The maximum area of the field is25 312.5 m2.

Problem 4 Max Area of Rectangle with Divisions

y = -2x + 450y = -2(112.5) + 450y = 225

3.2.7

112.5,25312.5V

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 0.7 100 1x x What does x represent?

Vertex (14.28, 9142.86)14 trees should be planted to maximize the revenue

AssignmentSuggested Questions

Page 174:1, 3, 5a,b, 7, 12, 15, 17, 23

3.2.8

Interpret a Graph: JavelinDave 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.

From ground level, what is the maximum height the javelin reached?

vertex (6, 19)

Maximum height is 19 m.

To reach maximum height, what 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.

Horizontal distance

height

From what height was the javelin thrown? 1 m3.2.5

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.

h = -5t2 + 100t

What is the maximum height reached by the flare?

vertex at (10, 500)

Maximum height is at 500 m.

How long is the flare in the air?

The flare is in the air for 20 seconds.

time

height

3.3A1.11