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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 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 = 253 Simply 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 = 121 x = 2 + √(121/9) or 2 - √(121/9) 9 9 x = 2 + [√(121) / √ (9)] or 2 – [√(121) / √(9)] (x – 2)² = 121/9 x = 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 c 6.25 + 26.01 = c² 2.5 32.26 = c² √(32.26) = c C B 5.68 = 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 - 12a 3x(4x + 5) – 5(4x + 5) • 3a² = 3a ∙ a The 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 + 10 x² - 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 - 6x x² + 15x • (½)(-6) = -3 (½)(15) = (15/2) • (-3)² → = 9 (15/2)² → = (15/2)² • The perfect-square The 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 = 0 2x² + 6x = 7 • + 24 +24 2(x² + 3x) = 7 • x² +10x = 24 x² + 3x = (7/2) • x²+10x+(5)²=24+(5)² x²+3x+(3/2)²=(7/2)+(3/2)² • x² + 10x + 25 = 49 x²+3x+(3/2)² =(7/2)+(9/4) • (x + 5)² = 49 [x + (3/2)]² = (23/4) • x + 5 = ± 7 x + (3/2) = ±√(23/4) • x = - 12 or x = 2 x = - (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.