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1 There are two types of solutions 1) Real solutions which are also x intercept(s) on the graph of the parabola b 2 4ac > 0 b 2 4ac = 0 2) Non real solutions which are not x intercept(s) on the graph of the parabola b 2 4ac < 0 Dec 8 12:38 PM 1

2 b2 4ac up or down which is considered a double solution Dec 8 12:38 PM 2

3 a=10, b= 13, c= 3 Check: Dec 8 12:44 PM 3

4 This is negative which means there are two non real solutions x= x= Dec 8 12:46 PM 4

5 a=15, b=2, c= 1 This is positive which means there are two real solutions x= Dec 8 12:46 PM 5

6 a=16, b= 72, c= =0 This is zero which means there are two indentical real solutions x= Dec 8 12:46 PM 6

7 a=1, b=7, c=1 This is positive which means there are two real solutions Dec 8 12:47 PM 7

8 50 =.005x 2 +2x =.005x 2 +2x 46.5 a =.005 b = 2, c = 46.5 Check: Dec 9 1:18 PM 8

9 2n+2 2n+4 2n (2n) 2 +(2n+2) 2 =(2n+4) 2 4n 2 +4n 2 +8n+4=4n 2 +16n+16 4n 2 8n 12=0 2n+2=2(3)+2=8 2n+4=2(3)+4=10 2n=2(3)=6 (2n) 2 +(2n+2) 2 =(2n+4) 2 Dec 9 1:21 PM 9

10 200 = 0.042s s = 0.042s s 196 a = 0.042, b = 1.1, c= 196 Dec 9 1:20 PM 10

11 a= 16, b=40, c= 45 Examine the value of the discriminant to determine how many real solutions exist 1600 (4)( 16)( 45) This is negative which means there are two non real solutions to the quadratic equation. This means the ball never reaches 50 feet. Dec 8 12:49 PM 11

12 Set y = 0 0 = 6x 2 + 5x 2 Examine the value of the discriminant to determine how many real solutions exist (5) 2 4(6)( 2) is positive so there are two real solutions indicating two x intercepts Dec 8 12:50 PM 12

13 Is the value of the discriminant a perfect square ( 1) 2 4(10)( 3) is a positive perfect square so there are two rational solutions Dec 8 12:50 PM 13

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15 0 = ax 2 +bx+c Mar 10 12:53 PM 15

16 Rewrite in vertex form: 1. y = x x 6 2. y = 2x 2 6x 1 3. y = 3x 2 30x + 12 Jul 30 6:30 PM 16

17 Quadratic formula If a x 2 + bx + c = 0 then x = b ± b 2 4ac 2a Sep 14 10:41 AM 17

18 Use the quadratic formula to solve 1. w 2 = 5w ( 3a + 2) ( 5a 1) = 2( 5a 1) Jul 30 6:32 PM 18

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46 Chapter 6 WS A 2,4,5,7.pdf Mar 11 1:20 PM 46

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54 Newton s Formula h(t) = 1 gt 2 + v o t + h o 2 Dec 12 9:09 PM 54

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95 Attachments Chapter 6 WS A 2,4,5,7.pdf Chapter 6 Lesson Masters B.pdf

AdvAlg6.4GraphingQuadratics.notebook. March 07, Newton s Formula h(t) = 1 gt 2 + v o t + h o 2. time. initial upward velocity

AdvAlg6.4GraphingQuadratics.notebook. March 07, Newton s Formula h(t) = 1 gt 2 + v o t + h o 2. time. initial upward velocity Notes Lesson 6 4 Applications of Quadratic Functions Newton s Formula h(t) = 1 gt 2 + v o t + h o 2 Height of object time Constant (accel. due to gravity) *32 ft/sec 2 *9.8 m/sec 2 **MEMORIZE THESE** initial

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