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.

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Section 5.1

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Section 5 1

Section 5.1

Introduction to Quadratic Functions


Quadratic function

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

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

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

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

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 values1

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

Section 5.2

Introduction to Solving Quadratic Equations


Solving equations of the form x a

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 x a1

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

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 roots1

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

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 theorem1

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

Section 5.3

Factoring Quadratic Expressions


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 expressions1

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 x bx c

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 x bx c1

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

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

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

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

Section 5.4

Completing the Square


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

Specific Case of a Perfect-Square Trinomial

  • x² + 8x + 16 = (x + 4)²

  • Understand: (½)8 = 4 → 4² = 16


Examples of completing the square

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

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

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 form1

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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