Graphing Quadratic and Higher Degree Polynomial Functions

is called a quadratic function. Let a, b, and c be real numbers a  0. The function f (x) = ax2 + bx + c is called a quadratic function. The graph of a quadratic function is a parabola. Every parabola is symmetrical about a line called the axis (of symmetry). x y The intersection point of the parabola and the axis is called the vertex of the parabola. f (x) = ax2 + bx + c vertex axis Quadratic function

The leading coefficient of ax2 + bx + c is a. y a > 0 opens upward When the leading coefficient is positive, the parabola opens upward and the vertex is a minimum. f(x) = ax2 + bx + c vertex minimum x y vertex maximum When the leading coefficient is negative, the parabola opens downward and the vertex is a maximum. f(x) = ax2 + bx + c a < 0 opens downward Leading Coefficient

Simple Quadratic Functions The simplest quadratic functions are of the form f (x) = ax2 (a  0) These are most easily graphed by comparing them with the graph of y = x2. Example: Compare the graphs of , and 5 y x -5 Simple Quadratic Functions

Example: Graph f (x) = (x – 3)2 + 2 and find the vertex and axis. f (x) = (x – 3)2 + 2 is the same shape as the graph of g (x) = (x – 3)2 shifted upwards two units. g (x) = (x – 3)2 is the same shape as y = x2 shifted to the right three units. - 4 x y 4 f (x) = (x – 3)2 + 2 g (x) = (x – 3)2 y = x 2 vertex (3, 2) Example: f(x) = (x –3)2 + 2

Quadratic Function in Standard Form The standard form for the equation of a quadratic function is: f (x) = a(x – h)2 + k (a  0) The graph is a parabola opening upward if a  0 and opening downward if a  0. The axis is x = h, and the vertex is (h, k). Example: Graph the parabola f (x) = 2x2 + 4x – 1 and find the axis and vertex. x y f (x) = 2x2 + 4x – 1 x = –1 f (x) = 2x2 + 4x – 1 original equation f (x) = 2( x2 + 2x) – 1 factor out 2 f (x) = 2( x2 + 2x + 1) – 1 – 2 complete the square f (x) = 2( x + 1)2 – 3 standard form a > 0  parabola opens upward like y = 2x2. (–1, –3) h = –1, k = –3  axis x = –1, vertex (–1, –3). Quadratic Function in Standard Form

Vertex and x-Intercepts Example: Graph and find the vertex and x-intercepts of f (x) = –x2 + 6x + 7. x y 4 f (x) = – x2 + 6x + 7 original equation (3, 16) x = 3 f (x) = – ( x2 – 6x) + 7 factor out –1 f (x) = – ( x2 – 6x + 9) + 7 + 9 complete the square f (x) = – ( x – 3)2 + 16 standard form a < 0  parabola opens downward. h = 3, k = 16  axis x = 3, vertex (3, 16). Find the x-intercepts by solving –x2 + 6x + 7 = 0. (7, 0) (–1, 0) (–x + 7 )( x + 1) = 0 factor x = 7, x = –1 x-intercepts (7, 0), (–1, 0) f(x) = –x2 + 6x + 7 Vertex and x-Intercepts

f (x) = a(x – h)2 + k standard form Example: Find an equation for the parabola with vertex (2, –1) passing through the point (0, 1). y x y = f(x) (0, 1) (2, –1) f (x) = a(x – h)2 + k standard form f (x) = a(x – 2)2 + (–1) vertex (2, –1) = (h, k) Since (0, 1) is a point on the parabola: f (0) = a(0 – 2)2 – 1 1 = 4a –1 and Example: Parabola

The vertex of the graph of f (x) = ax2 + bx + c (a  0) Vertex of a Parabola The vertex of the graph of f (x) = ax2 + bx + c (a  0) Example: Find the vertex of the graph of f (x) = x2 – 10x + 22. f (x) = x2 – 10x + 22 original equation a = 1, b = –10, c = 22 At the vertex, So, the vertex is (5, -3). Vertex of a Parabola

Example: Find the x-intercepts and vertex of the parabola f(x) = -x2 - 4x + 21

The maximum height of the ball is 15 feet. Example: A basketball is thrown from the free throw line from a height of six feet. What is the maximum height of the ball if the path of the ball is: The path is a parabola opening downward. The maximum height occurs at the vertex. At the vertex, So, the vertex is (9, 15). The maximum height of the ball is 15 feet. Example: Basketball

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: A fence is to be built to form a rectangular corral along the side of a barn 65 feet long. If 120 feet of fencing are available, what are the dimensions of the corral of maximum area? barn corral x 120 – 2x Let x represent the width of the corral and 120 – 2x the length. Area = A(x) = (120 – 2x) x = –2x2 + 120 x The graph is a parabola and opens downward. The maximum occurs at the vertex where a = –2 and b = 120 120 – 2x = 120 – 2(30) = 60 The maximum area occurs when the width is 30 feet and the length is 60 feet. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Maximum Area

A polynomial function is a function of the form where n is a nonnegative integer and each ai (i = 0, , n) is a real number. The polynomial function has a leading coefficient an and degree n. Examples: Find the leading coefficient and degree of each polynomial function. Polynomial Function Leading Coefficient Degree – 2 5 1 3 14 0 Polynomial Function

Power Functions

Exploration For each function, identify the degree and whether the degree is even or odd. Identify the leading coefficient and if it is positive or negative. Use a graphing calculator to graph each function. Describe the relationship between the degree and the sign of the leading coefficient with the right and left-hand behavior of the function. f(x) = x3 – 2x2 – x + 1 f(x) = 2x5 + 2x2 – 5x + 1 f(x) = -2x5 – x2 + 5x + 3 f(x) = -x3 + 5x – 2 f(x) = 2x2 + 3x – 4 f(x) = x4 – 3x2 + 2x – 1 f(x) = x2 + 3x + 2

The Leading Coefficient Test f(x) = anxn+… an > 0 an < 0 n even n odd

A polynomial function of degree n has at most n zeros. A real number a is a zero of a function y = f (x) if and only if f (a) = 0. Real Zeros of Polynomial Functions If y = f (x) is a polynomial function and a is a real number then the following statements are equivalent. 1. x = a is a zero of f. 2. x = a is a solution of the polynomial equation f (x) = 0. 3. (x – a) is a factor of the polynomial f (x). 4. (a, 0) is an x-intercept of the graph of y = f (x). A polynomial function of degree n has at most n zeros. Zeros of a Function

Example: Find all the real zeros of f (x) = x 4 – x3 – 2x2. Factor completely: f (x) = x 4 – x3 – 2x2 = x2(x + 1)(x – 2). y x –2 2 f (x) = x4 – x3 – 2x2 The real zeros are x = –1, x = 0, and x = 2. (–1, 0) (0, 0) These correspond to the x-intercepts. (2, 0) Example: Real Zeros

Example: Sketch the graph of f(x) = x3 – 2x2

Intermediate Value Theorem

Homework Page 134: 1-8, 19, 23, 25, 37, 39, 75, 76, 83 Page 148: 1-8, 13, 17, 21, 27, 33, 37, 39, 49, 57, 61, 67, 69, 75, 79, 90