1 / 36

Algebra 2 Chapter 5 Notes Quadratic Functions

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

obelia
Download Presentation

Algebra 2 Chapter 5 Notes Quadratic Functions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Algebra 2 Chapter 5 Notes Quadratic Functions

  2. Axis of Symmetry, The vertical line through the vertex 5.1 Graphing Quadratic Equations 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. 5.1 Vertex Form of a Quadratic Equations 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. 5.1 Intercept Form of a Quadratic Equation 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. 5.1 FOIL Method FOIL Method for changing intercept form or vertex form to standard form: [ First + Outer + Inner + Last ] ( x + 3 ) ( x + 5 ) = x2+ 5 x + 3 x + 15 = x2 + 8 x + 15

  7. 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 of28that 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 10and3 that combine to add up to − 17, when multiplied together?”

  8. 5.2 [ First + Outer + Inner + Last ] ( x + 3 ) ( x + 5 ) = x2+ 5 x + 3 x + 15 = x2 + 8 x + 15

  9. 5.2 Solving Quadratic Equations by Factoring

  10. 5.1 Solving Quadratic Equations 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. 5.1 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. 5.3 Solving Quadratic Equations Radical sign Radican: Radical ris 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 ± r x = x ½ 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. 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. 5.4 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 Complex Numbers 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. 5.4 Imaginary Number Imaginary Number i = − 1 and i2= − 1 Imaginary Number Squared

  16. 5.4 Imaginary Number What is the Square Root of – 25? ? = − 25 =− 1 25 =i ± 5

  17. 5.4 Complex Numbers ( Real number + imaginary number ) Complex Numbers (a + b i) 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. 5.4 Plot Complex Numbers Imaginary ( − 3 + 2 i ) ● Real ● (2 −3 i )

  19. 5.4 Complex Numbers: Add, Subtract, Multiply • ( 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. 5.4 Complex Numbers: Divide and Complex Conjugates CONJUGATEmeans 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. 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. 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. 5.1 Examples of Completing the Square 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. Completing the Square 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 ± i 3

  25. 5.1 Quadratic Functions in Vertex Form 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. Quadratic Formula 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  27. Quadratic Formula 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  28. Quadratic Formula 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  29. Quadratic Formula 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  30. Quadratic Formula 5.6 The Quadratic Formula and the Discriminantx= − b ± b2− 4 ac 2a

  31. Quadratic Formula 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. 5.6 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. 5.6 The Quadratic Formula and the Discriminant Using 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] 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. 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 ●

More Related