 # Algebra 2 Chapter 5 Notes Quadratic Functions.

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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 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

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

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

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

[ 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

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 )

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 ?

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

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

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 )

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

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 = –

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

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

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

( 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 )

5.4 Plot Complex Numbers Imaginary ( − i ) Real (2 − 3 i )

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

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 ]

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 │= − = ≈

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

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 ±

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

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 )

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

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

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

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

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

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

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

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

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

● ● 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