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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. 5.1. Graphing Quadratic Equations. 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 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 ●

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