1. Math 20-1 Chapter 3 Quadratic Functions 3.2 Quadratic Standard Form Teacher Notes

2. 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) = ax 2 + bx + c is called the ( standard form / vertex form ) of a quadratic function. Expressin standard form.

3. 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). y ≥ -4 3.2.2

5. 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 b) What is the value of the y-intercept? a) Graph the function. What does it represent? 25cm At time, t = 0 the height of the frog is 25cm. The from jumped from a height of 25cm. 3.2.3

6. 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

7. 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 -w 2 1000 = 2w + 2(length) 1000 – 2w = 2length length = 500 - w V(250, 62 500) Therefore, the dimensions of the lot are 250 m by 250 m. Problem 2 Max Area of Rectangle w = 250 3.2.5 w

8. Find the dimensions of the lot if only three sides need to be enclosed by the 1000 m of fence. Find the maximum area. xx 1000 - 2x For the length of fence: 1000 = 2x + length For the area: A = x(-2x + 1000) A = -2x 2 + 1000x V(250, 125 000) Therefore, the dimensions are 250 m x 500 m. The maximum area is 125 000 m 2. Problem 3 Max Area of Rectangle with Missing Side length = -2x + 1000 length = -2(250) +1000 length = 500 x = 250 3.2.6 Length = 1000 - 2x

9. 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. x xxx -2x + 450 length For the fence: 900 = 4x + 2length 450 = 2x + length length = -2x + 450 For the area: A = xlength A = x(-2x + 450) A= -2x 2 + 450x Therefore the dimensions of the Field are 112.5 m x 225 m. The maximum area of the field is 25 312.5 m 2. Problem 4 Max Area of Rectangle with Divisions y = -2x + 450 y = -2(112.5) + 450 y = 225 3.2.7

10. 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 = Maximum Revenue = What does x represent? Vertex (14.28, 9142.86) 14 trees should be planted to maximize the revenue

11. Page 174: 1, 3, 5a,b, 7, 12, 15, 17, 23 3.2.8

12. 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 = -0.5x 2 + 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 m 3.2.5

13. 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 = -5t 2 + 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