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Section 5.1

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Section 5.1

Introduction to Quadratic Functions

- A quadratic function is any function that can be written in the form f(x) = ax² + bx + c, where a ≠ 0.
- It is defined by a quadratic expression, which is an expression of the form as seen above.
- The stopping-distance function, given by: d(x) = ⅟₁₉x² + ¹¹̸₁₀x, is an example of a quadratic function.

- Let f(x) = (2x – 1)(3x + 5). Show that f represents a quadratic function. Identify a, b, and c.
- f(x) = (2x – 1)(3x + 5)
- f(x) = (2x – 1)3x + (2x – 1)5
- f(x) = 6x² - 3x + 10x – 5
- f(x) = 6x² + 7x – 5 a = 6, b = 7, c = - 5

- The graph of a quadratic function is called a parabola. Parabolas have an axis of symmetry, a line that divides the parabola into two parts that are mirror images of each other.
- The vertex of a parabola is either the lowest point on the graph or the highest point on the graph.

- The domain of any quadratic function is the set of all real numbers.
- The range is either the set of all real numbers greater than or equal to the minimum value of the function (when the graph opens up).
- The range is either the set of all real numbers less than or equal to the maximum value of the function (when the graph opens down).

- Let f(x) = ax² + bx + c, where a ≠ 0. The graph of f is a parabola.
- If a > 0, the parabola opens up and the vertex is the lowest point. The y-coordinate of the vertex is the minimum value of f.
- If a < 0, the parabola opens down and the vertex is the highest point. The y-coordinate of the vertex is the maximum value of f.

- f(x) = x² + x – 6
- Because a > 0, the parabola opens up and the function has a minimum value at the vertex.

- g(x) = 5 + 4x - x²
- Because a < 0, the parabola opens down and the function has a maximum value at the vertex.

Section 5.2

Introduction to Solving Quadratic Equations

- If x² = a and a ≥ 0, then x = √a or x = - √a, or simply x = ± √a.
- The positive square root of a, √a is called the principal square root of a.
- Simplify the radical for the exact answer.

- Solve 4x² + 13 = 253
- 4x² + 13 = 253Simply the Radical
- 13 - 13√60 = √(2 ∙ 2 ∙ 3 ∙ 5)

4x² = 240√60 = 2√(3 ∙ 5)

√60 = ± 2√15 (exact answer)

4x² = 240

4 4

x² = 60

x = √60 or x = - √60 (exact answer)

x = 7.75 or x = - 7.75 (approximate answer)

- Product Property of Square Roots:
- If a ≥ 0 and b ≥ 0: √(ab) = √a ∙ √b
- Quotient Property of Square Roots:
- If a ≥ 0 and b > 0: √(a/b) = √(a) ÷ √(b)

- Solve 9(x – 2)² = 121
- 9(x – 2)² = 121x = 2 + √(121/9) or 2 - √(121/9)
9 9

x = 2 + [√(121) / √ (9)] or 2 – [√(121) / √(9)]

(x – 2)² = 121/9x = 2 + (11/3) or 2 – (11/3)

√(x – 2)² = ±√(121/9)x = 17/3 or x = - 5/3

x – 2 = ±√(121/9)

x – 2 = √(121/9)

+ 2 + 2

- If ∆ABC is a right triangle with the right angle at C, then a² + b² = c²
A

c

a

C B

b

- If ∆ABC is a right triangle with the right angle at C, then a² + b² = c²
A

2.5² + 5.1² = c²

c6.25 + 26.01 = c²

2.532.26 = c²

√(32.26) = c

C B5.68 = c

5.1

Section 5.3

Factoring Quadratic Expressions

- When you learned to multiply two expressions like 2x and x + 3, you learned how to write a product as a sum.
- Factoring reverses the process, allowing you to write a sum as a product.
- To factor an expression containing two or more terms, factor out the greatest common factor (GCF) of the two expressions.

- 3a² - 12a3x(4x + 5) – 5(4x + 5)
- 3a² = 3a ∙ aThe GCF = 4x + 5
- 12a = 3a ∙ 4(3x – 5)(4x +5)
- The GCF = 3a
- 3a(a) – 3a ∙ 4
- (3a)(a – 4)

- To factor an expression of the form:
- ax² + bx + c where a = 1, look for integers r and s such that r ∙ s = c and r + s = b.
- Then factor the expression.
- x² + bx + c = (x + r)(x + s)

- x² + 7x + 10x² - 7x + 10
(5+2) = 7 & (5∙2) = 10 (-5-2) = -7 & (-5∙(-2)) = 10

(x + 5)(x + 2)(x – 5)(x – 2)

- a² - b² = (a + b)(a – b)
- (x + 3)(x – 3)
- x² + 3x - 3x - 9
- x² - 9
- x² - 3²

a² + 2ab + b² = (a + b)² a² - 2ab + b² = (a – b)²

(x + 3)² (x – 3)²

(x + 3)(x + 3) (x – 3)(x – 3)

x² + 3x + 3x + 9 x² - 3x – 3x + 9

X² +2(3x) + 9 x² - 2(3x) + 9

- If pq = 0, then p = 0 or q = 0.
- 2x² - 11x = 0
- x(2x – 11) = 0 (Factor out an x)
- x = 0 or 2x – 11 = 0
- x = 11 or x = 11/2

Section 5.4

Completing the Square

- When a quadratic equation does not contain a perfect square, you an create a perfect square in the equation by completing the square.
- Completing the square is a process by which you can force a quadratic expression to factor.

- x² + 8x + 16 = (x + 4)²
- Understand: (½)8 = 4 → 4² = 16

- x² - 6xx² + 15x
- (½)(-6) = -3 (½)(15) = (15/2)
- (-3)² → = 9(15/2)² → = (15/2)²
- The perfect-squareThe perfect square
- x² - 6x + 9 =x² + 15x + (15/2)² =
- (x – 3)²[x + (15/2)]²

- x² + 10x – 24 = 02x² + 6x = 7
- + 24 +242(x² + 3x) = 7
- x² +10x = 24x² + 3x = (7/2)
- x²+10x+(5)²=24+(5)²x²+3x+(3/2)²=(7/2)+(3/2)²
- x² + 10x + 25 = 49x²+3x+(3/2)² =(7/2)+(9/4)
- (x + 5)² = 49[x + (3/2)]² = (23/4)
- x + 5 = ± 7x + (3/2) = ±√(23/4)
- x = - 12 or x = 2x = - (3/2) + √(23/4) (0.90)
X = - (3/2) - √(23/4) (-3.90)

- If the coordinates of the vertex of the graph of y = ax² + bx + c, where a ≠ 0, are (h,k), then you can represent the parabola as:
- y = a(x – h)² + k, which is the vertex form of a quadratic function.

- Given g(x) = 2x² + 12x + 13
- 2(x² + 6x) + 13
- 2(x² + 6x + 9) + 13 – 2(9)
- 2(x + 3)² - 5
- 2[x – (-3)]² + (- 5) (Vertex Form)
- The coordinates (h,k) of the vertex are (-3, -5) and the equation for the axis of symmetry is x = - 3.