Quadratic Function Presentation

•? 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. OForms of Quadratic functions: * 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 = ? , 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.

. Using Quadratic Formula •? 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 The Graph of Quadratic Equation •? The graph of quadratic equation in the form f(x) = ax2+ bx+ c is a parabola. The parts of the graph of the parabola are determined by the values of a, b, and c.

•? The most meaningful points of the graph of a parabola are: 1. x-intercepts: The x-intercepts, if any, are also called the roots of the function. They are meaningful specifically as the zeroes of the function, but also represent the two roots for any value of . 2. -intercept: The importance of the y-intercept is usually as an initial value or initial condition for some state of an experiment, especially one where the independent variable represents time.

3. Vertex: The vertex represents the maximum (or minimum) value of the function, and is very important in calculus and many natural phenomena. X-intercepts •? •? •? •? •? The x intercepts of the graph of a quadratic function f given by y = ax2 + bx + c The x-intercepts are the solutions to the equation ax2 + bx + c = 0 The x-intercept in the equation f(x) = ax2 + bx + c, can be found in basically two ways, factoring or the quadratic formula. Factoring: If f(x) = y = ax2 + bx + c can be factored into the form y = a(x – r1)(x – r2) , then the x-intercepts are r1 and r2 . Example: y = x2 - 3x - 18 => 0 = x2 – 3x – 18 (Set y = 0 to find the x – intercept) 0 = (x – 6)(x + 3) (Factor) x – 6 = 0 or x + 3 = 0 x = 6 or x = -3 ( Solved the equation => x – int. = (6, 0) ; (-3, 0) ) * Quadratic formula: For any function in the form y = ax2 + bx + c, xintercepts are given by: ( , 0) Example: y = 2x2 + 5x + 3 (a = 2; b = 5; c = 3 ) => 0 = 2x2 + 5x + 3 (Set y = 0 to find x – int.

x = ( Using Quadratic formula to solve the equation) x = -5 ±1/4 => x = -19/4 or x = -21/4 (Simplified the equation => x – int. = (-21/4, 0) or (-19/4, 0)). •? •? Y-intercept •? The y intercept of the graph of a quadratic function is given by f(0) = c or y-intercept = (0,c). •? Example: Find the y intercept of the following equations: –? A) x2 + 2x + 26 –? B) 16x2 – 3x – 59 –? C) 10x2 + 5x – 10/7 * Solution: A) Substitute 0 for x as f(x) = f(0) = x2 + 2x + 26 => f(0) = c = 26.

The y intercept is at (0, 26). B) Same method as A). f(0) = -59. The y-int. s (0, -59) C) f(0) = -10/7.

The y-int. is (0, -10/7) •? The vertex can be found by completing the square, or by using the expression derived from completing the square on the general form. •? Formula given by: y = ax2 + bx + c So the vertex is •? Example: h = -12x2 + 168x – 38 •? => -b/2a = -168/2(-12) = 7 (applied the formula to find x-coordinate of vertex of the quadratic equation). •? Substitute for x (x = 7) into the original equation. We have that: h = -12(7)2 + 168(7) – 38 = 550 •? Thus, the vertex of the equation is (7, 550).

The Vertex Graphing Parabola ? There are several steps to do before we sketch a graph of parabola of the quadratic equation. •? The most easiest way to sketch the graph is from vertex form. •? Steps: 1. We need to check whether the graph is concave up if a > 0 (U- shaped) or concave down if a < 0(n-shaped).

2. We need to find the vertex of the equation by the formula x =-b/2a or by the vertex form. 3. We need to find the values of x and y intercepts. + Substitute x = 0 into the equation then find the coordinate of y value. + Let y = 0 then find the values of x.

Graphing Parabola •?Graph the function f(x) = 2x2 + 8x +7 Solution: 1.? Write the function f(x) = 2x2 + 8x + 7 in the form f(x) = a (x - h) 2 + k by completing the square. (h, k) f (x) = 2x2 + 8x + 7 => f (x) = 2 (x2 + 4x) + 7 => f (x) = 2 (x2 + 4x + 4) + 7 – 8 => f (x) = 2 (x + 2)2 – 1 3.? The function f (x) = 2x2 + 8x + 7 has vertex at (-2, -1) with a horizontal shift of 2 units to the left and a vertical shift of one unit downward. Also since a = 2 > 0 then the graph is concave up.

4.? The x-intercept of the function are determined by letting f (x) = 2 (x + 2)2 – 1 = 0 and solving for x as illustrated below:Continued f (x) = 2 (x + 2)2 – 1 = 0 => (x + 2)2 = ? => x + 2 = ± v? = ± 1/v2 => x = -2 ± (1/v2) 4. The y-intercept is f (0) = 2 (0 + 2)2 – 1 = 7 * The graph of f (x) = 2x2 + 8x +7 is illustrated below: Helpful Links •? h#p://www. douglas.

bc. ca/services/learning-­? centre/pdf/math/ MA7_30_QuadraBc_EquaBons_and_FuncBons . pdf   •? h#p://www. cimt.

plymouth. ac. uk/projects/ mepres/book9/y9s17ex. pdf   •? h#p://www. tech.

plym. ac. uk/maths/ resources/PDFLaTeX/quad_graphs. pdf   •? h#p://www. neufeldmath.

com/ss/Plus/ Graphing%209. pdf