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Section 5.1. Introduction to Quadratic Functions. Quadratic Function. 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.

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Section 5.1

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Section 5.1

Introduction to Quadratic Functions


Quadratic Function

  • 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.


Quadratic Functions

  • 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


Parabola

  • 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.


Domain and Range of Quadratic Functions

  • 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).


Minimum and Maximum Values

  • 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.


Minimum and Maximum Values

  • 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


Solving Equations of the Form x² = a

  • 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.


Solving Equations of the Form x² = a

  • 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)


Properties of Square Roots

  • 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)


Properties of Square Roots

  • 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


Pythagorean Theorem

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

    A

    c

    a

    C B

    b


Pythagorean Theorem

  • 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


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.


Factoring Quadratic 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)


Factoring x² + bx + c

  • 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)


Factoring x² + bx + c

  • 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)


Factoring the Difference of Two Squares

  • a² - b² = (a + b)(a – b)

  • (x + 3)(x – 3)

  • x² + 3x - 3x - 9

  • x² - 9

  • x² - 3²


Factoring Perfect-Square Trinomials

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


Zero-Product Property

  • 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


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.


Specific Case of a Perfect-Square Trinomial

  • x² + 8x + 16 = (x + 4)²

  • Understand: (½)8 = 4 → 4² = 16


Examples of Completing the Square

  • 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


Solving a Quadratic Equation by Completing the Square

  • 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)


Vertex Form

  • 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.


Vertex Form

  • 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.


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