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• A quadratic function (f) is a function that has the form as f(x) = ax2 + bx + c where a, b and c are real numbers and a not equal to zero (or a ≠ 0). • The graph of the quadratic function is called a parabola. It is a "U" or “n” shaped curve that may open up or down depending on the sign of coefficient a. Any equation that has 2 as the largest exponent of x is a quadratic function.

* Quadratic functions can be expressed in 3 forms: 1.  General form: f (x) = ax2 + bx + c 2.  Vertex form: f (x)= a(x - h)2 + k (where h and k are the x and y coordinates of the vertex) 3.  Factored form: f(x)= a(x - r1) (x - r2)

1. General form
•  Form : f(x) = ax2+ bx+ c •  General form is always written with the x2 term first, followed by the x term, and the constant term last. a, b, and c are called the coefficients of the equation. It is possible for the b and/or c coefficient to equal zero. Examples of some quadratic functions in standard form are: a. f(x) = 2x2 + 3x – 4 (where a = 2, b = 3, c = -4) b. f(x) =x2 – 4 (where a = 1, b = 0, c = -4) c. f(x)= x2 ( where a = 1, b and c = 0) d. f(x)= x2 – 8x (where a = ½, b = -8, c = 0).

2. Vertex Form
•  Form: f(x) = a(x - h)2 + k where the point (h, k) is the vertex of the parabola. •  Vertex form or graphing form of a parabola. •  Examples: a. 2(x - 2)2 + 5 (where a = 2, h = 2, and k = 5) b. (x + 5)2 (where a = 1, h = -5, k = 0)

3.Factored Form
•  Form: a(x - r1) (x - r2) where r1 and r2 are the roots of the equation. •  Examples: a. (x - 1)(x - 2) b. 2(x - 3)(x - 4) or (2x - 3)(x - 4)

Discriminant
•  Quadratic formula: If ax2 + bx + c, a ≠ 0, x = •  The value contained in the square root of the quadratic formula is called the discriminant, and is often represented by ∆ = b2 – 4ac. * b2 – 4ac > 0 → There are 2 roots x1,2= * b2 – 4ac = 0 * b2 – 4ac < 0 → → There is 1 real root, x = -b/2a. There are no real roots. .

•  A general formula for solving quadratic equations, known as the quadratic formula, is written as: •  To solve quadratic equations of the form ax2+ bx+ c, substitute the coefficients a, b, and c in to the quadratic formula. 1.  Exercise: Solve 4x2 – 5x + 1 = 0 using the quadratic formula ∆ = b2 – 4ac= 52 – 4(4)(1) = 25 – 16 = 9 => = =±3 x

x1 = 1

or

x2 = 1/4

Using Factoring
•  Convert from general form, f(x) = ax2 + bx + c to factored form, a(x - x1) (x - x2) : + Example 1: Solve x2+ 2x = 15 by factoring => x2 + 2x – 15 = 0 (General form) (x + 5)(x - 3) = 0 (Factoring form) x+5=0 or x–3=0 x = -5 or x=3 → Thus, the solution to the quadratic equation is x = -5 or x = 3. + Example 2: Solve x2 + 5x – 9 = -3 by factoring => x2 + 5x – 6 = 0 (General form) (x – 1)(x + 6) = 0 (Factoring form) x–1=0 or x+6=0 x=1 or x = -6 → Thus, the solution to the quadratic equation is x = 1 or x = -6.

Using Completing the Square
•  Converting from the general form f(x) = ax2+ bx+ c to a statement of the vertex form f(x)= a(x - h)2 + k. •  When quadratic equations cannot be solved by factoring, they can be solved by the method of completing the squares. •  Example: Solve x2 + 4x – 26 = 0 by completing the square •  (x2 + 4x + 4 – 4) – 26 = 0 (General form) •  (x + 2)2 – 4 – 26 = 0 •  (x + 2)2 – 30 = 0 (Vertex form) •  (x + 2)2 = 30 •  x + 2 = •  x =-2 •  x =-2+ or x = - 2

Application to higher-degree equations
•  Example: x4 + 4x2 - 5 = 0 –  The equation above can be written as: –  (x2)2 + 4(x2) - 5 = 0 –  (Quadratic function with exponent x = 2) –  Solve: ./ Substitute x2 = P P2 + 4P - 5 =0 (P + 5)(P - 1) = 0 P = -5 or P=1 ./ Re-substitute P = x2 => P = -5 = x2 = no roots => P = 1 = x2 => x = ± 1