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Algebra 2 Chapter 5 Notes Quadratic Functions

Algebra 2 Chapter 5 Notes Quadratic Functions. Axis of Symmetry , The vertical line through the vertex. NOTES: Page 56, Section 5.1 Graphing a Quadratic Function Quadratic Function in standard form: y = a x 2 + b x + c Quadratic functions are U-shaped, called “ Parabola .”. ●.

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Algebra 2 Chapter 5 Notes Quadratic Functions

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  1. Algebra 2 Chapter 5 Notes Quadratic Functions

  2. Axis of Symmetry, The vertical line through the vertex NOTES: Page 56, Section 5.1 Graphing a Quadratic Function Quadratic Function in standard form: y = ax2 + bx + c Quadratic functions are U-shaped, called “Parabola.” ● Graph of a Quadratic Function: If parabola opens up, then a > 0 [POSITIVE VALUE] If parabola opens down, then a < 0 [NEGATIVE VALUE] 2. Graph is wider than y = x2 , if│a│< 1 Graph is narrower than y = x2 , if │a│> 1 3. x-coordinate of vertex = ─ b 2 a 4. Axis of symmetry is one vertical line, x = ─ b 2 a Vertex, Lowest or highest point of the quadratic function Example: Graph y = 2 x2– 8 x + 6 a = 2 , b = ─ 8 , c = 6 Since a > 0 , parabola opens up

  3. NOTES: Page 57, Section 5.1 Vertex & Intercept Forms of a Quadratic Function Vertex form: y = a ( x – h )2 + k (− 3 , 4 ) ● Example 1: Graph y = −1 ( x + 3 )2 + 4 2 ● ● (− 5 , 2 ) (− 1 , 2 ) x =− 3

  4. NOTES: Page 57a, Section 5.1 Vertex & Intercept Forms of a Quadratic Function Intercept form: y = a ( x –p)( x – q ) (1 , 9 ) ● Example 2: Graph y = −1 ( x + 2) ( x – 4) (− 2 , 0 ) (4 , 0 ) ● ● x =1 y = −1 ( 1 + 2) ( 1– 4) Y = 9

  5. Graphing Quadratic Equations

  6. NOTES: Page 57b, Section 5.1 FOIL Method for changing intercept form or vertex form to standard form: [ First + Outter + Inner + Last ] ( x + 3 ) ( x + 5 ) = x2+ 5 x + 3 x + 15 = x2 + 8 x + 15

  7. NOTES: Page 58, Section 5.2 Solving Quadratic Equations by Factoring Use factoring to write a trinomial as a product of binomials x2 + b x + c = ( x + m ) ( x + n ) = x2 + ( m + n ) x + m n So, the sum of ( m + n ) must = b and the product of m n must = c Example 1 : Factoring a trinomial of the form, x2 + b x + c Factor: x2 − 12 x − 28 “What are the factors of 28that combine to make a difference of − 12?” Example 2 : Factoring a trinomial of the form, ax2 + b x + c Factor: 3x2 − 17 x + 10 “What are the factors of 10and 3 that combine to add up to − 17, when multiplied together?”

  8. [ First + Outter + Inner + Last ] ( x + 3 ) ( x + 5 ) = x2+ 5 x + 3 x + 15 = x2 + 8 x + 15

  9. NOTES: Page 58a, Section 5.2 Solving Quadratic Equations by Factoring

  10. NOTES: Page 59, Section 5.2 Zero Product Property: If A • B = 0, the A = 0 or B= 0 With the standard form of a quadratic equation written as ax2 + bx + c = 0, if you factor the left side, you can solve the equation.

  11. NOTES: Page 59a, Section 5.2 Finding Zeros of Quadratic Functions x – intercepts of the Intercept Form: y = a (x – p ) ( x – q) p = (p , 0 ) and q = (q , 0) Example: y = x2 – x – 6 y = ( x + 2 ) ( x – 3 ), then Zeros of the function are p = – 2 and q = 3. • • (– 2 , 0 ) ( 3 , 0 )

  12. NOTES: Page 60, Section 5.3 Solving Quadratic Functions r is a square root of s if r2 = s 3 is a square root of 9 if 32 = 9 Since (3)2 = 9 and (-3)2 = 9, then 2 square roots of 9 are: 3 and – 3 Therefore, ± or ± Radical sign Radican: Radical x = x ½ r 3 r 9 Square Root of a number means: What # times itself = the Square Root of a number? Example: 3 • 3 = 9, so the Square Root of 9 is 3. Product Property ab = a • 36 4 • 9 b = ( a > 0 , b > 0) Quotient Property a b 4 9 a 4 = = b 9 Examples: = 6 = 2 6 24 4 = = 90 • = 10 6 • 15 9 10 3

  13. NOTES: Page 60a, Section 5.3 Solving Quadratic Functions A Square Root expression is considered simplified if No radican has a Perfect Square other than 1 There is no radical in the denominator Examples “Rationalizing the denominator” 7 14 2 7 7 7 2 7 16 = = = = 4 16 2 2 2 Solve: 2 x2 + 1 = 17 2 x2 = 16 x2 = 8 X = ± 4 X = Solve: 1 3 ( x + 5)2 = 7 ( x + 5)2 = 21 ( x + 5)2 = 21 x + 5 = 21 X = – 5 21 ± • ± 2 2 X = – 5 21 ± 2 ± + { and ± X = – 5 21 –

  14. NOTES: Page 61, Section 5.4 Complex Numbers Because the square of any real number can never be negative, mathematicians had to create an expanded system of numbers for negative number Called theImaginary Unit “ i “ Defined as i = − 1 and i2 = − 1 Property of the square root of a negative number: If r = + real number, then − r = − 1 • r = − 1 • r = i r − 5 = − 1 • 5 = − 1 • 5 = i 5 ( i r )2 = − 1 • r = − r ( i 5 )2 = − 1 • 5 = − 5 Solving Quadratic Equation 3 x2 + 10 = − 26 3 x2 = − 36 x2 = − 12 x2 = − 12 x = − 12 x = − 1 12 x = i 4 • 3 x = ± 2 i 3

  15. Imaginary Number i = − 1 and i2 = − 1 Imaginary Number Squared

  16. What is the Square Root of – 25? ? = − 25 =− 1 25 =i ± 5

  17. NOTES: Page 61a, Section 5.4 Complex Numbers (a + b i) ( Real number + imaginary number ) Imaginary Numbers Real Numbers ( a+b i) ( a+0 i) ( 2+3 i) ( 5−5 i) − 1 5 2 Pure Imaginary Numbers 3 ( 0 + b i ) , where b ≠ 0 ∏ 2 ( − 4 i) ( 6 i)

  18. NOTES: Page 62, Section 5.4 Complex Numbers: Plot Imaginary ( − 3 + 2 i ) ● Real ● (2 −3 i )

  19. NOTES: Page 62a, Section 5.4 Complex Numbers: Add and Subtract • ( 4− i ) + ( 3− 2 i ) = 7− 3 i • ( 7− 5 i ) − ( 1− 5 i ) = 6 + 0 i • 6 − ( − 2 + 9 i) +(− 8 + 4 i)=0− 5 i= − 5 i Complex Numbers: Multiply a) 5 i ( − 2 + i ) = − 10 i + 5 i2= − 10 i+ 5 ( − 1) =− 5 − 10 i b) ( 7−4 i) ( − 1 +2 i ) = b) ( 6+ 3 i) ( 6− 3 i ) = − 7 + 4 i+ 14 i− 8 i2 36+ 18 i− 18 i− 9 i2 − 7 + 18 i− 8 (−1) − 7 + 18 i+ 8 1+ 18 i 36 + 0 i− 9 (−1) 36 + 0 + 9 45

  20. NOTES: Page 62b, Section 5.4 Complex Numbers: Divide and Complex Conjugates CONJUGATE means to “Multipy by same real # and same imaginary # but with opposite sign to eliminate the imaginary #.” 5 + 3 i 1− 2 i 1 + 2 i 1+ 2 i 5+ 10 i + 3 i+ 6 i2 1+ 2 i – 2 i– 4 i2 = • 5+ 13 i + 6 (– 1 ) 1 – 4 (– 1 ) = – 1 + 13 i 5 = – 1 + 13 i 5 5 [ standard form ] =

  21. NOTES: Page 62, Section 5.4 Complex Numbers: Absolute Value Imaginary ( − 1 + 5 i ) Z =a + b i │ Z │ =a2 + b2 ● (3 +4 i ) ● Absolute Value of a complex number is a non-negative real number. Real ( −2 i ) ● │ 3 + 4 i│ = 32 + 42 = 25 = 5 │ −2 i│ = │ 0 + 2 i│ = 02 + ( − 2 )2 = 2 c) │− 1 + 5 i│= − 12 + 52 = 26 ≈ 5.10

  22. NOTES: Page 64, Section 5.5 Completing the Square b 2 x b x x x b x 2 x2 bx x2 b 2 b 2 b x 2 ( b )2 ( 2 ) RULE: x2 + b x + c, where c = ( ½ b )2 In a quadratic equation of a perfect square trinomial, the Constant Term = ( ½ linear coefficient ) SQUARED. x2 + b x + ( ½ b )2 = ( x + ½ b )2 Perfect Square Trinomial = the Square of a Binomial

  23. Example 1 x2 − 7 x + c “What is ½ of the linear coefficient SQUARED?” c= [ ½ (− 7 ) ] 2 = ( − 7 ) 2 = 49 2 4 x2 − 7 x + 49 4 = ( x − 7)2 2 Perfect Square Trinomial = the Square of a Binomial Example 2 x2 + 10 x − 3 “Is − 3half of the linear coefficient SQUARED?” [ if NOT then move the − 3 over to the other side of = , then replace it with the number that is half of the linear coefficient SQUARED ] c= [ ½ (+ 10 ) ] 2 = ( 5 ) 2 = 25 x2 + 10 x − 3 = 0 x2 + 10 x =+ 3 x2 + 10 x + 25 =+ 3 + 25 ( x + 5 )2 = 28 ( x + 5 )2 = 28 x + 5 = 4 7 x = – 5 ± 2 7

  24. NOTES: Page 65, Section 5.5 Completing the Square where the coefficient of x2 is NOT “ 1 “ 3 x2 – 6 x + 12 = 0 3 x2 – 6 x + 12 = 0 3 x2 – 2 x + 4 = 0 As + 4 isn’t [ ½ (– 2) ]2 , move 4 to other side of = x2 – 2 x = – 4 x2 – 2 x + 1 = – 4 + 1 What is [ ½ (– 2) ]2= (– 1)2= 1 ? ( x – 1 )2 = – 3 ( x – 1 )2 = – 3 ( x – 1 ) = – 1 3 x = + 1 ± i3

  25. NOTES: Page 65a, Section 5.5 Writing Quadratic Functions in Vertex Form y = a ( x − h)2 + k y = x2 – 8 x + 11 11 doesn’t work here, so move 11 out of the way and replace the constant “c” with a # that makes a perfect square trinomial y + 16 = ( x2 – 8 x + 16 ) + 11 What is [ ½ ( – 4 ) ]2 = (– 4 )2 = 16 y + 16 = ( x – 4 )2 + 11 ( x2 – 8 x + 16 ) = ( x – 4 )2 – 16 – 16 y = ( x – 4 )2 – 5 ( x , y ) = ( 4 , – 5 )

  26. NOTES: Page 66, Section 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  27. NOTES: Page 66a, Section 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  28. NOTES: Page 67, Section 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  29. NOTES: Page 67, Section 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  30. NOTES: Page 67, Section 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  31. NOTES: Page 68, Section 5.6 The Quadratic Formula and the Discriminant

  32. Quadratic Equation in Standard Form: a x2 + b x + c = 0 3 x2 – 11 x – 4 = 0 x= − b ± b2− 4 ac 2a x= + 11 ± (11)2− 4 (3) (– 4) 2 (3) x= + 11 ± 121+ 482 (3) x= + 11 ± 1696 x= + 11 ± 13 = 24 , – 2 = 4 , – 16 6 6 3 Sum of Roots: – b a 4 + – 1 = 11 3 3 Product of Roots: c a 4 • – 1 = – 4 3 3

  33. Solve this Quadratic Equation: a x2 + b x + c = 0 x2 + 2 x – 15 = 0 Factoring Quadratic Formula x 2 + 2 x – 15 = 0 ( x – 3 ) ( x + 5 ) = 0 x – 3 = 0 or x + 5 = 0 x = 3 or x = – 5 x= − b ± b2− 4 ac 2a x 2 + 2 x – 15 = 0 x= – 2 ± (– 2)2− 4 (1) (– 15) 2 (1) x= – 2 ± 4 + 602 x= – 2 ± 642 x= – 2 ± 8= 3 or – 5 2 Completing the Square x 2 + 2 x – 15 = 0 x 2 + 2 x = + 15 x 2 + 2 x + 1 = + 15 + 1 ( x + 1 ) 2 = 16 ( x + 1 ) = 16 x = – 1 ± 4 = 3 or – 5

  34. NOTES: Page 68a, Section 5.6 The Quadratic Formula and the Discriminant IMMAGINARY x2 − 6 x + 10 = 0 = 3 ± i No intercept x2 − 6 x + 9 = 0 = 3 One intercept Two intercepts x2 − 6 x + 8 = 0 = 2or4 REAL ● ● ●

  35. y > a x2 + b x + c [graph of the line is a dash] y ≥ a x2 + b x + c [graph of the line is solid] y < a x2 + b x + c [graph of the line is a dash] y ≤ a x2 + b x + c [graph of the line is solid] NOTES: Page 69, Section 5.7 Graphing & Solving Quadratic Inequalities Vertex (standard form) = − b = − (−2 ) = 1 2a 2 (1 ) y = 1 x2−2 x − 3 y = 1 (1)2−2 (1) − 3 = − 4 Vertex = ( 1 , − 4 ) Line of symmetry = 1 Example 1: y > 1 x2−2 x − 3 0 = (x − 3 ) ( x + 1 ) So, either (x − 3 ) = 0 or ( x + 1 ) = 0 Then x = 3 or x = − 1 ● ● ● Test Point (1,0) to determine which side to shade y > 1 x2−2 x − 3 0 > 1 (1)2−2 (1) − 3 0 > 1 −2 − 3 0 > − 4 This test point is valid, so graph this side ●

  36. NOTES: Page 69a, Section 5.7 Graphing & Solving Quadratic Inequalities y ● x ● ● ● ● y < − x2 − x + 2 y < − ( x2 + x − 2 ) y < − ( x − 1 ) ( x + 2 ) y < − x2 − x + 2 y < − ( − 1)2 − (− 1 ) + 2 2 2 y < − 1 + 1 + 2 4 2 y < 2 1 4 y ≥ x2 − 4 y ≥ ( x − 2 ) ( x + 2 ) x = −b = 0 = 0 2a 2 y ≥ x2− 4 y ≥ (0)2 − 4 y ≥ − 4 ●

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