1. Chapter 4 Applications of Quadratic Models

2. To graph the quadratic equation y = ax 2 + bx +c  Use vertex formula x v = -b/2a  Find the y-coordinate of the vertex by substituting x, into the equation of parabola  Locate x-intercepts by setting y= 0  Locate y-intercept by evaluating y for x = 0  Locate axis of symmetry  Vertex form for a Quadratic Formula where the vertex of the graph is x v, y v y = a(x – x v ) 2 + y v

3. The x- coordinate of the vertex of the graph of y = ax 2 + bx x v = -b/2a y = 2x 2 + 8x + 6 x v = - 8/2(2) Substitute – 2 for x = - 2 y v = 2(-2) + 8( -2) + 6 = 8 – 16 + 6 = -2 So the vertex is the point (-2, -2) The x-intercepts of the graph by setting y equal to zero 0 = 2x 2 + 8x + 6 = 2(x + 1)(x + 3) x + 1 = 0 or x + 3 = 0 x = -1, x = -3 The x-intercepts are the points (-1, 0) and (-3, 0) And y-intercept = 6 (-2, -2) -3 (-2, -8) - 5 6

4. Problem 39, Pg 211 a b). The price of a room is 20 + 2x, the number of rooms rented is 60 – 3x The total revenue earned at that price is (20 + 2x) (60 – 3x) c). Enter Y1 = 20 + 2x Y2 = 60 – 3x Y3 = (20 + 2x)(60 – 3x) in your calculator Tbl start = 0 Tb1 = 1 The values in the calculator’s table should match with table d). If x = 20, the total revenue is 0 e). Graph f). The owner must charge atleast $24 but no more than $36 per room to make a revenue atleast $1296 per night g). The maximum revenue from night is $1350, which is obtained by charging $30 per room and renting 45 rooms at this price a 200 1500

5. x be the no of price increases Price of room = 20 + 2x No. of rooms rented = 60 – 3x Total revenue = (20 + 2x)(60 – 3x) 0 20 60 1200 1 22 57 1254 2 24 54 1296 3 26 51 1326 4 28 48 1344 5 30 45 1350 6 32 42 1344 7 34 39 1326 8 36 36 1296 10 40 30 1200 12 44 24 1056 16 52 12 624 20 60 0 0 No of price Price of room No. of rooms rented Total revenue increases Max. Revenue Lowest Highest

6. 4.2 Using Calculator for Quadratic Regression Pg 216, Ex 4 Graph STAT Enter datas Press Y = and select Plot 1 then press ZOOM 9 Store in Y1by pressing STAT right 5 VARS right 1, 1 Enter

7. 4.2 Ex 13, Page 221 0 2000 4000 Cable 500 Tower 20 Tower The vertex is (2000, 20) and another point on the cable is (0, 500). Using vertex form, y = a(x – 2000) 2 + 20 Use point (0, 500) 500 = a(0 – 2000) + 20, 500 = 4,000,000a + 20 480 = 4,000,000a a = 0.000012 The shape of the cable is given by the equation y = 0.00012(x – 2000) 2 + 20

8. 4.3 Maximum and Minimum Values Example 1 a) Revenue = (price of one item) (number of items sold) R = x(600 – 15x) R = 600x – 15x 2 b) Graph is a parabola c) x v = - b/2a = -600/2(-15) = 20 R v= 600(20) – 15(20) 2 = 6000 20 40 6000 5000 R = 600x – 15x 2 Late Nite Blues should charge $20 for a pair of jeans in order to maximize revenue at $6000 a week

9. Solving Systems with the Graphing Calculator Example 2 Page 225 Enter Y1= Enter Window Press 2 nd, table press graph Enter Y1, Y2 Press window Press 2 nd and calc Press graph Pg 226

10. 4.4 Quadratic Inequalities Example 1, Pg 235 h = 256 t – 16t 2 256 t – 16 t 2 > 800 1000 800 500 h = 256 t – 16t 2 5 10 15 t h The solution set 4.25 < t < 11.75 Compound inequality

11. Interval Notation An interval is a set that consists of all the real numbers between two numbers a and b. If the set includes both of the end points a and b, so that a< x < b, then the set is called a closed interval Denoted by [a, b] If the set does not include its endpoints, so that a<x<b, then it is called an open interval, and is denoted by (a, b) -2 -1 0 1 2 3 x 2 (-, -2 ) U ( 2, )

12. Examples Half open or half closed interval 3 < x < 6 0 1 2 3 4 5 6 7 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 [ -9, ) - 1 0 1 2 3 4 5 6 ( -, 1] U ( 4, ) x > 9 x 4

13. * - 8 -1 0 2 a.y > 0 for –8 < x < 2. Using interval notation [-8, 2] b.y 2. Using interval notation ( -, -8) U ( - 1, ) c.y -1. Using interval notation (, -5) U (- 1, ) d.y > 21 for –5 < x < -2. Using interval notation (-5, -2) Y1 =16 – 6x - x^2 Y2 = 21 X min = -9.4 Xmax = 9.4 Ymin = -25 Ymax = 25 4.4 Ex 20, Pg 241