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 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 xy(x,y) Evaluate the following equation for the given domain and then determine if it represents a function. y = 2x - 2x D={-1, 0, 1, 2} 2x - 2x ( ) -2( ) (-1,3) (0,-1) (1,-1) (2,3) 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 yx(x-axis,y-axis) Evaluate the following equation for the given domain and then determine if it represents a function. x = 2y - 2y D={-1, 0, 1, 2} 2y - 2y ( ) -2( ) (3,-1) (-1,0) (-1,1) (3,2) 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 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 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 Quadratic term Lineal term Constant term y = 4x +3x +8 2

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 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 y = ( ) + 4( ) – 2 y = 4 – y = – 3 Vertex: (– 2, – 3) Axis of symmetry: Finding a point to the right of the vertex: Using x = 0 y = ( ) + 4( ) y = 1 ( 0, 1) x y x= -2 (– 2, – 3) ( 0, 1 )

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 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved Graph the following using vertex form of the quadratic function: y = x + 4x y = (x + 4x + )+ 1 – (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( ) y = 1 ( 0, 1) x y x= -2 (– 2, – 3) ( 0, 1 )

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 – (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 – (-1)(8) = +7 (-2)(4) = +2 (x – 2)(x + 4) = 0 x – 2 = 0 x + 4 = 0 +2 x = 2 -4 x = -4 (2,0) (-4,0) x y x= -1 (– 1, – 9) ( 2, 0 ) ( -4, 0 )

11 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved y = x – 4 x y = (x – 4 x + ) + 3 – (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:

x= -( ) ( ) - 4( )( ) 2( ) 2 +_ 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 = x= 2 +_ 4 2 x= 2 +_ 4 2 x= x= 2 _ x = 3 x = 1 LETS GO BACK TO THE PROBLEM!

13 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved y = x – 4 x y = (x – 4 x + ) + 3 – (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 +_ 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 Standard 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved y = x + 2x – 4 2 y = (x + 2x + ) – 4 – (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 = x + 2x + = (1) 2 2 (x + 1) = 5 2 x + 1 = 5 +_ x = 5 – 1 +_ _ 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 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: x y (6,0) (0,0) 8 (3, – 3) = a(-3) – = a(9) – = 9a 9 9 a= 1 3 Using: (0,0)(0,0) (x,y) y= (x – 3) + (– 3) y= (x – 3) – y = a(x-h) + k 2

16 Standard 4 Find the parabola that passes through points (1,0), (7,0), ( 10,9). If Y = aX +bX +c 2 (x, y ) and using the given points. ( ) = a( ) + b( ) + c = 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 = a= = 48a + 6b Using IV: 0 = 48( ) + 6b = b = 6b 6 6 b = Using I: 0 = a + b + c 1 3 – c = 0 – 7 3 c= 7 3 Using II and III:

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