1. 1 QUADRATIC FUNCTIONS PROBLEM 1 PROBLEM 2 PROBLEM 3 INTRODUCTION Standard 4, 9, 17 PROBLEM 4 PROBLEM 5 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved PROBLEM 6 PROBLEM 7

2. 2 STANDARD 4: Students factor polynomials representing the difference of squares, perfect square trinomials, and the sum and difference of two cubes ESTÁNDAR 4: Los estudiantes factorizan polinomios representando diferencia de cuadrados, trinomios cuadrados perfectos, y la suma de diferencia de cubos. ALGEBRA II STANDARDS THIS LESSON AIMS: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 3 xy(x,y) 0 1 2 Evaluate the following equation for the given domain and then determine if it represents a function. y = 2x - 2x - 1 2 D={-1, 0, 1, 2} 2x - 2x - 1 2 2 2( ) -2( )- 1 2 2 2 0 0 1 1 2 2 3 3 (-1,3) (0,-1) (1,-1) (2,3) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y We can guess that is a parabola and observe that all over the x values, it has only one corresponding y value. So IT IS A FUNCTION. We can also perform the vertical line test and verify that it is a function because it crosses the curve a only one point. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved REVIEW

4. 4 yx(x-axis,y-axis) 0 1 2 Evaluate the following equation for the given domain and then determine if it represents a function. x = 2y - 2y - 1 2 D={-1, 0, 1, 2} 2y - 2y - 1 2 2 2( ) -2( )- 1 2 2 2 0 0 1 1 2 2 3 3 (3,-1) (-1,0) (-1,1) (3,2) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y We can guess that is a parabola that opens to the right but most values in the domain have two values in the range. So it is not a FUNCTION. We can also perform the vertical line test and verify that it is NOT a function because it crosses the curve at MORE THAN ONE POINT. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved REVIEW

5. 5 Standard 4 Features of the graph of a quadratic function (parabola): axis of symmetry x y vertex PRESENTATION CREATED BY SIMON PEREZ. All rights reserved root zero solution

6. 6 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved Identify: Quadratic, Lineal and Constant terms. Quadratic term Lineal term Constant term y = 2x - 2x - 1 2 Quadratic term Lineal term Constant term y = 4x +3x +8 2

7. 7 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved Graph the following using the graph features of the quadratic function: y = x + 4x + 1 2 y = ax + bx + c 2 a = 1 b = 4 c = 1 Finding the axis of symmetry: x = – b 2a2a – ( ) 2( ) 4 1 x = – 4 2 x = – 2 Finding the vertex: y = x + 4x + 1 2 y = ( ) + 4( ) + 1 2 – 2 y = 4 – 8 + 1 y = – 3 Vertex: (– 2, – 3) Axis of symmetry: Finding a point to the right of the vertex: Using x = 0 y = ( ) + 4( ) + 1 2 00 y = 1 ( 0, 1) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y x= -2 (– 2, – 3) ( 0, 1 )

8. 8 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved Upward if a>0 Downward if a<0 Direction of opening (h,k)Vertex x= hAxis of symmetry y = a(x-h) + kForm of equation VERTEX FORM EQUATION 2

9. 9 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved Graph the following using vertex form of the quadratic function: y = x + 4x + 1 2 y = (x + 4x + )+ 1 – 2 4 2 2 4 2 2 2 (2) 2 2 y = (x + 4x + )+ 1 – 2 4 (4) y = (x + 2) + 1 – 4 2 y = (x + 2) – 3 2 Rewriting the equation: y = (x - -2) + (-3) 2 y = a(x-h) + k 2 h= -2 k= -3 a= 1 Vertex: (h,k) = (-2,-3) Axis of symmetry:x= -2 Finding a point to the right of the vertex: Using x = 0 y = ( ) + 4( ) + 1 2 00 y = 1 ( 0, 1) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y x= -2 (– 2, – 3) ( 0, 1 )

10. 10 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved Graph the following function by finding the vertex, axis of symmetry and roots: y = x + 2x – 8 2 y = (x + 2x + ) – 8 – 2 2 2 2 2 2 2 2 (1) 2 2 y = (x + 2x + ) – 8 – 2 1 (1) y = (x + 1) – 8 – 1 2 y = (x + 1) – 9 2 Rewriting the equation: y = (x - -1) + (-9) 2 y = a(x-h) + k 2 h= -1 k= -9 a= 1 Vertex: (h,k) = (-1,-9) Axis of symmetry: x= -1 Finding the roots by factoring: x + 2x – 8 = 0 2 – 8 + 2 (-1)(8) -1+ 8 = +7 (-2)(4) -2+ 4 = +2 (x – 2)(x + 4) = 0 x – 2 = 0 x + 4 = 0 +2 x = 2 -4 x = -4 (2,0) (-4,0) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y x= -1 (– 1, – 9) ( 2, 0 ) ( -4, 0 )

11. 11 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved y = x – 4 x + 3 2 y = (x – 4 x + ) + 3 – 2 4 2 2 4 2 2 2 (2) 2 2 y = (x – 4 x + ) + 3 – 2 4 (4) y = (x – 2) + 3 – 4 2 y = (x – 2) – 1 2 Rewriting the equation: y = (x – 2) + (-1) 2 y = a(x-h) + k 2 h= 2 k= -1 a= 1 Vertex: (h,k) = (2,-1) Axis of symmetry: x= 2 Finding the roots by Quadratic Formula: x – 4 x + 3 = 0 2 X= -b b - 4ac 2a2a 2 +_ where:0 = aX +bX +c 2 a = 1 b = - 4 c = 3 Graph the following function by finding the vertex, axis of symmetry and roots:

12. x= -( ) ( ) - 4( )( ) 2( ) 2 +_ 1 1 -4 3 x= 4 16 – ( 4 )( 3 ) 2 +_ 4 16 – 12 x= 2 +_ PRESENTATION CREATED BY SIMON PEREZ. All rights reserved Finding the roots by Quadratic Formula: x – 4 x + 3 = 0 2 X= -b b - 4ac 2a2a 2 +_ where:0 = aX +bX +c 2 a = 1 b = - 4 c = 3 4 4 x= 2 +_ 4 2 x= 2 +_ 4 2 x= 2 + 4 2 x= 2 _ x = 3 x = 1 LETS GO BACK TO THE PROBLEM!

13. 13 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved y = x – 4 x + 3 2 y = (x – 4 x + ) + 3 – 2 4 2 2 4 2 2 2 (2) 2 2 y = (x – 4 x + ) + 3 – 2 4 (4) y = (x – 2) + 3 – 4 2 y = (x – 2) – 1 2 Rewriting the equation: y = (x – 2) + (-1) 2 y = a(x-h) + k 2 h= 2 k= -1 a= 1 Vertex: (h,k) = (2,-1) Axis of symmetry: x= 2 Finding the roots by Quadratic Formula: x – 4 x + 3 = 0 2 a = 1 b = - 4 c = 3 x= -( ) ( ) - 4( )( ) 2( ) 2 +_ 1 1 -4 3 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y x= 2 (2, – 1) ( 3, 0 ) ( 1, 0 ) x = 3 x = 1 (1,0) (3,0) Graph the following function by finding the vertex, axis of symmetry and roots:

14. 14 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved y = x + 2x – 4 2 y = (x + 2x + ) – 4 – 2 2 2 2 2 2 2 2 (1) 2 2 y = (x + 2x + ) – 4 – 2 1 (1) y = (x + 1) – 4 – 1 2 y = (x + 1) –5 2 Rewriting the equation: y = (x - -1) + (-5) 2 y = a(x-h) + k 2 h= -1 k= -5 a= 1 Vertex: (h,k) = (-1,-5) Axis of symmetry: x= -1 Finding the roots by completing the square: x + 2x – 4 = 0 2 +4 x + 2x + = 4 + 2 2 2 2 2 2 2 2 (1) 2 2 (x + 1) = 5 2 x + 1 = 5 +_ x = 5 – 1 +_ _ 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y x= -1 (– 1, – 5) (1.2, 0) (-3.2, 0) Graph the following function by finding the vertex, axis of symmetry and roots: x = 1.2 x = -3.2

15. 15 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved ( ) = a(( ) – ( )) + ( ) 2 y = a(x-h) + k 2 h= 3 k= -3 Vertex: (h,k) = (3,-3) Find the equation of the graph below: 1 2 -2 3 - 3 4 -4 5 -5 6-6 7 -7 1 2 -2 3 -3 4 -4 5 -5 x y (6,0) (0,0) 8 (3, – 3) 0 0 3 -3 0 = a(-3) – 3 2 0 = a(9) – 3 +3 3 = 9a 9 9 a= 1 3 Using: (0,0)(0,0) (x,y) y= (x – 3) + (– 3) 2 1 3 y= (x – 3) – 3 2 1 3 y = a(x-h) + k 2

16. 16 Standard 4 Find the parabola that passes through points (1,0), (7,0), ( 10,9). If Y = aX +bX +c 2 (x, y ) 1 1 2 2 3 3 and using the given points. ( ) = a( ) + b( ) + c 2 2 2 1 1 77 10 0 0 9 0 = a + b + c 0 = 49a + 7b + c 9 = 100a + 10b + c Solving the equations together: 0 = a + b + c 0 = 49a + 7b + c 9 = 100a + 10b + c IIIIII Using I and II: 0 = a + b + c 0 = 49a + 7b + c (-1) 0 = -a – b – c 0 = 49a + 7b + c 0 = 48a + 6b IV (-1) 0 = 49a + 7b + c 9 = 100a + 10b + c 0 = -49a – 7b – c 9 = 100a + 10b + c 9 = 51a + 3 b V Solving IV and V: (-2) 0 = 48a + 6b 9 = 51a + 3 b 0 = 48a + 6b -18 = -102a – 6b -54a = -18 -54 a= 1 3 0 = 48a + 6b Using IV: 0 = 48( ) + 6b 1 3 0 = 16 + 6b -16 -16 = 6b 6 6 b = - 8 3 Using I: 0 = a + b + c 1 3 – 8 3 + c = 0 – 7 3 c= 7 3 Using II and III:

17. 17 Standard 4 Find the parabola that passes through points (1,0), (7,0), ( 10,9). If Y = aX +bX +c 2 (x, y ) 1 1 2 2 3 3 and using the given points. ( ) = a( ) + b( ) + c 2 2 2 1 1 77 10 0 0 9 0 = a + b + c 0 = 49a + 7b + c 9 = 100a + 10b + c Solving the equations together: 0 = a + b + c 0 = 49a + 7b + c 9 = 100a + 10b + c b = - 8 3 c= 7 3 a= 1 3 Y = aX +bX +c 2 Then: Y = X X + 2 – 8 3 7 3 1 3 Y = (X – 8X + 7) 2 1 3 Y = (X – 8X + ___ + 7 – ___ ) 2 1 3 8 2 2 8 2 2 (4) 2 2 16 Y = (X – 8X + ___ + 7 – ___ ) 2 1 3 2 1 3 Y = ((X – 4) – 9) 2 1 3 Y = (X – 4) – 3 2 1 3