Download presentation

1
Algebra 2 Chapter 5 Notes Quadratic Functions

2
**Graphing Quadratic Equations**

Axis of Symmetry, The vertical line through the vertex 5.1 Graphing Quadratic Equations Quadratic Function in standard form: y = a x2 + b x + 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 ● 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 X- coordinate = ─ b 2 a ─ (─8) 2 (2) = 8 4 = 2 } Vertex ( x , y ) ( 2 , ─ 2 ) Y- coordinate 2 (2)2 – 8 (2) + 6 ─ 2

3
**Vertex Form of a Quadratic Equations**

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 a = − 1 2 Since a < 0 , parabola opens down h − 3 k 4 Vertex ( h , k ) (− 3 , 4 ) Axis of symmetry : x = − 3 Plot 2 pts on one side of axis of symmetry ● ● (− 5 , 2 ) (− 1 , 2 ) x = − 3

4
**Intercept Form of a Quadratic Equation**

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 ) a = − 1 Since a > 0 , parabola opens down p − 2 q 4 X –intercepts = ( 4 , 0 ) and (− 2 , 0 ) Axis of symmetry : x = 1 , which is halfway between the x-intercepts Plot 2 pts on one side of axis of symmetry (− 2 , 0 ) (4 , 0 ) ● ● x = 1 y = −1 ( ) ( 1 – 4 ) Y = 9

5
**Graphing Quadratic Equations**

Name of Form Equation Form How do you find the x-coordinate of the Vertex Standard y = ax2 + bx + c x = – b 2 Then substitute x into equation to get y of the vertex, then substitute another value for x to get another point Vertex y = a (x – h) 2 + k Vertex is (h, k) then substitute another value for x to get another point Intercept y = a (x – p) (x – q) x = midpoint between p and q, then substitute x into equation to get y of the vertex, then substitute another value for x to get another point

6
**[ First + Outer + Inner + Last ]**

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
**Solving Quadratic Equations by Factoring**

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 28 that combine to make a difference of − 12?” Factors of = ( 28 • 1 ) ( 14 • 2 ) ( 7 • 4 ) Example 2 : Factoring a trinomial of the form, ax2 + b x + c Factor: 3x2 − 17 x + 10 “What are the factors of 10 and 3 that combine to add up to − 17, when multiplied together?” Factors of = ( 10 • 1 ) ( 5 • 2 ) 3 • − • − 2 = − 17 Factors of 3 = ( 3 • 1 )

8
**5.2 [ First + Outer + Inner + Last ] ( x + 3 ) ( x + 5 )**

= x2 + 5 x + 3 x + 15 = x2 + 8 x + 15 Signs of Binomial Factors for Quadratic Trinomials The four possibilities when the quadratic term is + ax2 + bx + c ( ) ( ) x2 + 8x + 15 ( x + 5 ) ( x + 3 ) What are the factors of 15 that add to + 8 ? ax2 – bx + c ( – ) ( – ) y2 – 7y + 12 ( y – 4 ) ( y – 3 ) What are the factors of 12 that add to – 7 ? ax2 + bx – c ( ) ( – ) where + is > r2 + 7r – 18 ( r + 9 ) ( r – 2 ) What are the factors of 18 that have difference of + 7 ? ax2 – bx – c ( – ) ( ) where – is > z2 – 6z – 27 ( z – 9 ) ( z + 3 ) What are the factors of 27 that have difference of – 6 ?

9
**Solving Quadratic Equations by Factoring Special Factoring Patterns**

5.2 Solving Quadratic Equations by Factoring Special Factoring Patterns Name of pattern Pattern Example Difference of 2 Squares a2 – b2 = ( a + b ) ( a – b ) x2 – 9 = ( x + 3) ( x – 3 ) Perfect Square Trinomial { a2 + 2ab + b2 = ( a + b ) 2 x x + 36 = ( x + 6 ) 2 a2 – 2ab + b2 = ( a – b ) 2 x2 – 8x + 16 = ( x – 4 ) 2 4 x2 – 25 = (2x) 2 – (5) 2 = (2 x + 5 ) (2 x – 5) Difference of 2 Squares 9 y y + 16 = (3y) (3y)(4) + 42 = (3y + 4) 2 Perfect Square Trinomial 49 r2 – 14r + 1 = (7r) 2 – 2 (7r) (1) = ( 7r – 1) 2

10
**Solving Quadratic Equations Factoring Monomials First**

5.1 Solving Quadratic Equations Factoring Monomials First 5x2 – 20 = 5 ( x2 – 4) = 5 (x + 2) (x – 2) 6p2 – 15p + 9 = 3 (2p2 – 5 p + 3) = 3 ( 2p – 3) ( p – 1 ) 2u2 + 8 u = 2 u ( u + 4) 4 x2 + 4x + 4 = 4 ( x2 + x + 1) 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. Solve Quadratic Equations x2 + 3x – 18 = 0 (x – 3) (x + 6) = 0 x – 3 = 0 or x + 6 = 0 x = 3 or x = – 6 2t2 – 17t = 3t – 5 2t2 – 20t + 50 = 0 t2 – 10t = 0 (t – 5) 2 = 0 t – 5 = 0 t = 5

11
**Finding Zeros of Quadratic Functions**

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
**Solving Quadratic Equations**

5.3 Solving Quadratic Equations Radical sign Radican: Radical 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 ± r x = x ½ 3 r 9 Examples of Perfect Squares 4 is 2x2 9 is 3x3 16 is 4x4 25 is 5x5 36 is 6x6 49 is 7x7 64 is 8x8 81 is 9x9 100 is 10x10 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 • b 36 = 4 • 9 ( a > 0 , b > 0) Quotient Property a b a 4 9 4 = = b 9 Examples: 24 = 4 6 = 2 6 = = 6 • 15 90 9 • 10 = 3 10

13
**Solving Quadratic Functions “Rationalizing the denominator”**

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 2 14 7 7 7 16 7 2 = = = = 4 16 2 2 2 Solve: 2 x2 + 1 = 17 2 x2 = 16 x2 = X = ± X = Solve: 1 3 ( x + 5)2 = 7 ( x + 5)2 = 21 ( x + 5)2 = x + 5 = x = – • 2 ± 2 2 x = – + { and x = – –

14
**5.4 Complex Numbers − 5 = − 1 • 5 = − 1 • 5 = i 5**

Because the square of any real number can never be negative, mathematicians had to create an expanded system of numbers for negative number Called the Imaginary Unit “ i “ Defined as i = − 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 • = i ( i r )2 = − 1 • r = − r ( i )2 = − 1 • 5 = − 5 Solving Quadratic Equation 3 x = − 26 3 x2 = − 36 x2 = − 12 x2 = − 12 x = − 12 x = − x = i • 3 x = ± 2 i 3

15
**Imaginary Number Squared**

5.4 Imaginary Number Imaginary Number i = − 1 and i2 = − 1 Imaginary Number Squared

16
**What is the Square Root of – 25?**

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

17
**( Real number + imaginary number ) Pure Imaginary Numbers**

5.4 Complex Numbers ( Real number + imaginary number ) Complex Numbers ( a + b i ) Imaginary Numbers Real Numbers ( a + b i ) ( a + 0 i ) ( 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 ( − i ) ● Real ● (2 − 3 i )

19
**Complex Numbers: Add, Subtract, Multiply**

5.4 Complex Numbers: Add, Subtract, Multiply ( 4 − i ) + ( 3 − 2 i ) = 7 − 3 i ( 7 − 5 i ) − ( 1 − 5 i ) = i 6 − ( − i ) + ( − 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 ) ( − i ) = b) ( i ) ( 6 − 3 i ) = − i i − 8 i 2 i − 18 i − 9 i 2 − i − 8 (−1) − i + 8 i i − 9 (−1) 45

20
**Complex Numbers: Divide and Complex Conjugates**

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 #.” Complex Conjugates ( a + b i ) • ( a − b i ) = REAL # ( i ) • ( 6 − 3 i ) = REAL # i 1 − 2 i i i + 3 i + 6 i2 i – 2 i – 4 i2 = • i + 6 (– 1 ) 1 – 4 (– 1 ) = – i 5 = – i = [ standard form ]

21
**Complex Numbers: Absolute Value**

5.4 Complex Numbers: Absolute Value Imaginary ( − i ) Z = a + b i │ Z │ = a2 + b2 ● ( i ) ● Absolute Value of a complex number is a non-negative real number. Real ● ( − 2 i ) │ i │ = = = 5 │ − 2 i │ = │ i │ = ( − 2 )2 = 2 c) │− i │= − = ≈

22
**5.5 Completing the Square RULE: x2 + b x + c, where c = ( ½ 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
**Examples of Completing the Square**

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 x2 − 7 x + 49 4 = ( x − 7 )2 2 Perfect Square Trinomial = the Square of a Binomial Example 2 x x − 3 “Is − 3 half 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 x x − 3 = 0 x x = + 3 x x = ( x + 5 )2 = 28 ( x + 5 )2 = x = x = – 5 ±

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 – 2 x = As + 4 isn’t [ ½ (– 2) ]2 , move 4 to other side of = x2 – 2 x = – 4 x2 – 2 x = – What is [ ½ (– 2) ]2 = (– 1)2 = 1 ? ( x – 1 )2 = – 3 ( x – 1 ) = – x = ± i

25
**Quadratic Functions in Vertex Form**

5.1 Quadratic Functions in Vertex Form Writing Quadratic Functions in Vertex Form y = a ( x − h )2 + k y = x2 – 8 x 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 = ( x2 – 8 x + 16 ) What is [ ½ ( – 4 ) ]2 = (– 4 )2 = 16 y = ( x – 4 ) ( x2 – 8 x + 16 ) = ( x – 4 )2 – – 16 y = ( x – 4 )2 – 5 ( x , y ) = ( 4 , – 5 )

26
Quadratic Formula 5.6 The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a Quadratic Equation a x2 + b x + c = 0 Divide by a to both sides of = x2 + b x + c = 0 a a − c to both sides a x2 + b x = − c a a Complete the square ( + to both sides of = ) x2 + b x + ( b )2 = − c + ( b )2 = b2 − 4 a c [ combine both terms] a ( 2a ) a ( 2a ) a2 Binomial Squared (x + b )2 = b2 − 4 a c 2a a2 Square Root both sides of = (x + b )2 = b2 − 4 a c 2a a2 Squared Root undoes Squared x + b = ± b2 − 4 a c 2a a Solve for x by − b to both sides 2a x = − b ± b2 − 4 a c 2a a = − b ± b2 − 4 a c

27
Quadratic Formula 5.6 The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a Number and type of solutions of a quadratic equation determined by the DISCRIMINANT If b2 − 4 a c > 0 Then equation has 2 real solutions If b2 − 4 a c = 0 Then equation has 1 real solutions If b2 − 4 a c < 0 Then equation has 2 imaginary solutions

28
**Ex 1: Solving a quadratic equation with 2 real solutions**

Quadratic Formula 5.6 The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a Ex 1: Solving a quadratic equation with 2 real solutions a x2 + b x + c = 0 x = − b ± b2 − 4 a c 2 a 2 x2 + 1 x = 5 x = − 1 ± − 4 (2 ) ( −5 ) 2 ( 2 ) 2 x2 + 1 x − 5 = 0 x = − 1 ± 4

29
**Ex 2: Solving a quadratic equation with 1 real solutions**

Quadratic Formula 5.6 The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a Ex 2: Solving a quadratic equation with 1 real solutions a x2 + b x + c = 0 x = − b ± b2 − 4 a c 2 a 1 x2 − 1 x = 5 x − 9 x = − (−6) ± (−6) 2 − 4 (1 ) ( 9) 2 ( 1 ) 1 x2 − 6 x + 9 = 0 x = ± 2 x = 3

30
**Ex 3: Solving a quadratic equation with 2 imaginary solutions**

Quadratic Formula 5.6 The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a Ex 3: Solving a quadratic equation with 2 imaginary solutions a x2 + b x + c = 0 x = − b ± b2 − 4 a c 2 a −1 x2 + 2 x = 2 x = − (2) ± (2) 2 − 4 (−1 ) (− 2) 2 (−1 ) −1 x2 + 2 x − 2 = 0 x = − (2) ± − 4 2 x = − 2 ± 2 i −2 x = − 2(1 ± I ) x = 1 ± i

31
**Quadratic Formula 5.6 The Quadratic Formula and the Discriminant**

EQUATION DISCRIMINANT SOLUTIONS a x2 + b x + c = 0 b2 − 4 a c x = − b ± b2 − 4 a c 2 a x2 − 6 x + 10 = 0 (− 6 )2 − 4 (1) (10 ) = − 4 x = − (− 6 ) ± − 4 2 (1) x = − (− 6 ) ± 2 i = 3 ± i x2 − 6 x + 9 = 0 (− 6 )2 − 4 (1) (9) = 0 x = − (− 6 ) ± x = − (− 6 ) ± = 3 x2 − 6 x + 8 = 0 (− 6 )2 − 4 (1) (8) = 4 x = − (− 6 ) ± x = − (− 6 ) ± = 2 or 4

32
**x = − b ± b2 − 4 ac 2a a x2 + b x + c = 0 3 x2 – 11 x – 4 = 0**

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)2 − 4 (3) (– 4) (3) x = ± (3) x = ± x = ± = 24 , – 2 = 4 , – Sum of Roots: – b a 4 + – 1 = Product of Roots: c a 4 • – 1 = – 4

33
**5.6 Factoring Quadratic Formula Completing the Square**

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

34
**Using the Discriminant**

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 = 2 or 4 REAL ● ● ●

35
**● ● Graphing & Solving Quadratic Inequalities**

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] Vertex (standard form) = − b = − (− 2 ) = 1 2a (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 x Y 1 − 4 3 ● ● ● 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
**Graphing & Solving Quadratic Inequalities**

5.7 Graphing & Solving Quadratic Inequalities y x y − 1 2 2 1 4 − 2 1 x y − 4 2 −2 ● x ● ● ● ● y < − x2 − x + 2 y < − ( x2 + x − 2 ) y < − ( x − 1 ) ( x + 2 ) y < − ( − 1 )2 − (− 1 ) + 2 y < − y < 2 1 4 y ≥ x2 − 4 y ≥ ( x − 2 ) ( x + 2 ) x = −b = 0 = 0 2a y ≥ ( 0 )2 − 4 y ≥ − 4 ●

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google