Polynomial Functions Polynomial Functions An equation having a

Polynomial Functions

Polynomial Functions �An equation having a variable raised to an exponent of a non negative integer �Degree: the highest power on the variable second degree �i. e. f(x) = x 2 + 1 third degree f(x) = x 3 + 1 first degree f(x) = x + 2 zero degree (no variable) f(x) = 3

Zero Degree Polynomials Constant Functions

Constant Functions �A constant function is described by the rule f(x) = b where b can be any number �The rate of change (slope) is zero �Horizontal line with equation y = b (0, 3) y = 3

First Degree Polynomials Linear Functions

Linear Functions �A linear function is described by the rule f(x) = ax + b where a cannot equal 0 �The parameter “b” is the y-intercept �The parameter “a” represents the rate of change (slope) �If a > 0, rate of change is positive �If a < 0, rate of change is negative Rate of Change = y 2 – y 1 x 2 – x 1 Point 1 (x 1, y 1) Point 2 ( x 2, y 2) 2 -2 2

Second Degree Polynomials Quadratic Functions

Quadratic Function Standard Form f(x) = a(x-h)2 + k a>0 General Form Vertex: (h, k) Axis of symmetry: x = h f(x) = ax 2 + bx + c Zeros: h Vertex: Factored Form f(x) = a(x-x 1)(x-x 2) Zeros: x 1 and x 2 Axis of Symmetry: x = a<0 Axis of symmetry: x = Zeros: y intercept: c

Example The following formula links the speed v in km/h and the distance d in metres necessary to bring a car to a complete stop in case of emergency. 100 d = v 2 + 20 v At what speed is a car travelling if it needs 63 m to come to a complete stop? 100(63) = v 2 + 20 v 6300 = v 2 + 20 v - 6300

Example The parabola represented below crosses the x-axis at the points (-1, 0) and (3, 0) and its vertex is the point P(1, ‑ 4). f(x) = a(x-h)2 + k f(x) = a(x-1)2 - 4 0= a(3 -1)2 - 4 0= 4 a - 4 a=1 f(x) = (x-1)2 - 4 f(x) = (x-1) - 4 f(x) = x 2 - x – x +1 - 4 y = x 2 2 x 3 What is the equation of the parabola graphed above in general form?

Example �One leg of a right triangle exceeds the other leg by four inches. The hypotenuse is 20 inches. Find the length of the longer leg of the right triangle. c 2 = a 2 + b 2 20 x x+4 202 = x 2 + (x + 4) 2 400 = x 2 + (x + 4) 400 = x 2 + (x 2 + 8 x + 16) 0 = 2 x 2 + 8 x -384

Example The height, h(t), in feet of an object above the ground is given by h(t) = -16 t 2 + 64 t + 190, where t is the time in seconds. a) Find the time it takes the object to strike the ground b) Find the maximum height of the object. Vertex:

Example A diver jumps off the side of an inground swimming pool and reaches a depth of 2. 6 meters at a distance of 1. 95 m away from the ledge. He enters the water at a distance of 1. 76 meters from the ledge. Determine the maximum height during his trajectory assuming that he followed a parabolic motion. Vertex: f(x) = a(x-x 1)(x-x 2) f(x) = a(x-0)(x-1. 76) -2. 6 = a(1. 95 -0)(1. 95 -1. 76) -2. 6 = a(1. 95)(0. 19) -2. 6 = 0. 3705 a a = -7. 02 (1. 76, 0) f(x) = -7. 02 x(x-1. 76) f(x) = -7. 02 x 2 + 12. 36 x (1. 95, -2. 6)