1. Material on Quiz and Exam Student will be able to:  If given Quadratic Function in Standard Form:  ID Vertex, Axis of Symmetry, x and y intercepts  Sketch Parabola  Rewrite equation into Quadratic Fcn‘s Std Form  ID LH and RH Behavior of Polynomials  Perform Long Division of Polynomials  Perform Synthetic Division of Polynomials

2. f(x)=a(x-h) 2 +k h  Axis of symmetry, its a vertical line at x=h k  Vertex at ( h, k ). Note: ‘h’ is (must be) subtracted from x. also a)If a>0 then it opens upwards, if a<0, then it opens downwards. b)‘a’ also tells us the “fatness”/ “skinniness” of the parabola

3. Substitute O for x (y intercept) and O for y (x intercept) e.g. f(x)= 3(x-1) 2 - 9 0= 3(x – 1) 2 – 9  9=3(x – 1) 2  9/3=(x – 1) 2  3=(x – 1) 2  ±√3 = x – 1  (±√3 +1, 0) x intercept e.g. f(x)= 3(x-1) 2 - 9 f(x)= 3(0-1) 2 - 9 f(x)= 3( – 1) 2 – 9  f(x)= 3 – 9  f(x)= – 6  ( 0, – 6) y intercept

4. using h, k and intercepts f(x)=5(x-3) 2 + 4 f(x)=1(x+3) 2 – 4 Then graph these

6. We can write its function, f(x) e.g., if h=3, and k=5 and the parabola passes thru (0,0)… f(x)=a(x-h) 2 +k f(x)=a(x-3) 2 +5 We can find a by substitution … 0=a(0-3) 2 +5 –5=9a – 5/9=a f(x)= – 5/9(x-3) 2 +5

7.  The steps to sketching a parabola: 1. Put in Standard Form 2. Identify the Vertex 3. Determine Direction 4. Determine where it crosses the y-axis 5. Sketch it. 1. f(x)=a(x-h)2+k  ____________ 2.Vertex  ( ___, ___ ) 3.Direction  _________ 4.Y intercept  ( ____, ____ )

8.  ID the vertex, the direction of the parabola & its y-intercept: a) f(x) = 5(x – 2) 2 + 4 b) f(x) = –5(x – 1) 2 + 3 c) f(x) = – 2(x + 3) 2 + 4 d) f(x) = (x – 3) 2 + 2 Vertex  ( ___, ___ ) Direction  ___________ Y intercept  ( ____, ____ )

9.  Now Graph each of them. a) f(x) = 5(x – 2) 2 + 4 b) f(x) = –5(x – 1) 2 + 3 c) f(x) = – 2(x + 3) 2 + 4 d) f(x) = (x – 3) 2 + 2

10. Put in Std Form by Completing the Square!! When ‘a’ = 1  a(x 2 +bx)+c Coefficient in front of x 2 must be 1. take half of the ‘b” term and square it, (b/2) 2. Add the result to the expression inside the parenthesis above and subtract it as well. a(x 2 +bx+ (b/2) 2 ) + c − (b/2) 2 This creates a perfect square trinomial  (x+b/2) 2 + ck o Putting into standard Quadratic Function gives: − o (x – − b/2) 2 + c, h= − b/2 and k=c Becomes the c in the standard formula

11. Put in Std Form by Completing the Square When ‘a’ not equal to 1 ! Coefficient in front of x 2 must be 1, so pull the ‘a’ out  a(x 2 +bx)+c take half of the ‘b” term and square it, (b/2) 2. Add the result to the expression inside the parenthesis and subtract it as well  a(x 2 +bx+ (b/2) 2 − (b/2) 2 ) + c Simplify to create a perfect square trinomial  (x+b/2) 2 + c − a(b/2) 2 o Putting into standard Quadratic Function gives: − o (x – − b/2) 2 + c, h= − b/2 and k=c Becomes the c in the standard formula

12. Put in Std Form by Completing the Square When ‘a’ not equal to 1 ! Coefficient in front of x 2 must be 1, so pull the ‘a’ out  a(x 2 +bx)+c take half of the ‘b” term and square it, (b/2) 2. Add the result to the expression inside the parenthesis and subtract it as well  a(x 2 +bx+ (b/2) 2 − (b/2) 2 ) + c Simplify to create a perfect square trinomial  (x+b/2) 2 + c − a(b/2) 2 o Putting into standard Quadratic Function gives: − o (x – − b/2) 2 + c, h= − b/2 and k=c Becomes the c in the standard formula

13. page 272, #s 78-84 Groups of four with these roles: o Solver – math problem solution leader o Recorder – easel pad leader o Speaker during Gallery Walk o Time Keeper Time tables are on the white board ! So is the format of the expected easel chart !

14. Simple Example… from class 1x 2 – 2x 1 [ 1x 2 –2x +(2/2) 2 – (2/2) 2 ] + 0 1[x 2 –2x +(1) 2 ] – 1 (1) 2 + 0 [x–1] 2 – 1

15. Example… Problem #78 –.008x 2 +1.8x + 1.5 –.008 [ 1x 2 –225x +(112.5) 2 – (112.5) 2 ] + 1.5 –.008[x 2 –225x +(112.5) 2 ] + (–.008) (– (112.5) 2 + 1.5 –.008[x–112.5] 2 + (–.008)(– (112.5) 2 + 1.5 –.008[x–112.5] 2 + 102.75

16. OR Memorize for ax 2 +bx+c, memorize: Vertex is [ –b/2a, f(–b/2a) ] Axis of symmetry is at x= –b/2a