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Graphing Quadratic and Higher Degree Polynomial Functions
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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
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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
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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
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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
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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 – original equation f (x) = 2( x2 + 2x) – factor out 2 f (x) = 2( x2 + 2x + 1) – 1 – 2 complete the square f (x) = 2( x + 1)2 – 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
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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 original equation (3, 16) x = 3 f (x) = – ( x2 – 6x) factor out –1 f (x) = – ( x2 – 6x + 9) complete the square f (x) = – ( x – 3) 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) = factor x = 7, x = –1 x-intercepts (7, 0), (–1, 0) f(x) = –x2 + 6x + 7 Vertex and x-Intercepts
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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
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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 original equation a = 1, b = –10, c = 22 At the vertex, So, the vertex is (5, -3). Vertex of a Parabola
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Example: Find the x-intercepts and vertex of the parabola f(x) = -x2 - 4x + 21
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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
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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 = –2x 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
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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 – Polynomial Function
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Power Functions
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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
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The Leading Coefficient Test
f(x) = anxn+… an > 0 an < 0 n even n odd
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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
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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
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Example: Sketch the graph of f(x) = x3 – 2x2
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Intermediate Value Theorem
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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
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