3 Polynomial and Rational Functions © 2008 Pearson Addison-Wesley. All rights reserved Sections 3.1–3.2

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Quadratic Functions and Models 3.2 Synthetic Division Polynomial and Rational Functions 3

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-3 Quadratic Functions and Models 3.1 Quadratic Functions ▪ Graphing Techniques ▪ Completing the Square ▪ The Vertex Formula ▪ Quadratic Models and Curve Fitting

Polynomial Functions A polynomial function of degree n, where n is a nonnegative integer, is a function defined by an expression of the form where are real numbers where. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-4

Quadratic Function A function f is a quadratic function if where a, b, and c are real numbers with a ≠ 0. The graph of a quadratic function is a parabola. Parabolas have a line of symmetry thru the vertex. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-5

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 1(b) Graphing Quadratic Functions (page 304) Graph g(x) = –2x 2 and compare to y = x 2 and y = 2x 2. Give the domain and range. The graph of g(x) = –2x 2 is a narrower version of the graph of y = x 2 and is a reflection of the graph of y = 2x 2 across the x-axis. Domain: Range:

Graphing Techniques for the Quadratic Equation Standard form of the quadratic function: 1.The vertex is the point (h,k). The axis of symmetry is x = h. 2.The orientation is given by a : –if a > 0 the graph opens up –if a < 0 the graph opens down 4.The y-intercept is f(0) = c. 5.The x-intercept(s) are the solutions to f(x) = 0. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-7

Graphing Techniques for the Quadratic Equation Non-Standard form of the quadratic function: 1.The vertex is the point. The axis of symmetry is x = -b/2a. 2.The orientation is given by a : –if a > 0 the graph opens up –if a < 0 the graph opens down 4.The y-intercept is (0,c). 5.The x-intercept(s) are the solutions to f(x) = 0. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-8

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-9 Graphing Quadratic Function in standard form Graph F(x) = –2(x + 3) using transformations. Give the domain and range. The graph of F(x) is the graph of g(x) translated 3 units to the left and 5 units up. It is stretched vertically by a factor of 2. it is reflected across the x-axis. Domain: Range:

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Graphing a Parabola with the non-standard form Graph f(x) = x 2 + 2x – 5 by locating the vertex and the x-and y-intercepts. The graph opens up because a > 0. The vertex is found by: The y-intercept is (0,-5) The x-intercepts are the solutions to. This calls for the quadratic formula. Vertex (–1, –6) Axis: x = –1

Graphing a Parabola with the non-standard form Copyright © 2008 Pearson Addison-Wesley. All rights reserved The two x-intercepts are:

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 2 – Non-standard form for Quadratic Function Find the axis and vertex of the parabola f(x) = –3x x – 8 using the vertex formula. Axis: x = 2 a = –3, b = 12, c = –8 Vertex: (2, f(2)) Vertex: (2, 4)

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 5(a) Solving a Problem Involving Projectile Motion (page 308) A ball is thrown directly upward from an initial height of 75 ft with an initial velocity of 112 ft per sec. Give the function that describes the height of the ball in terms of time t. The projectile height function is

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 5(b) Solving a Problem Involving Projectile Motion (page 308) After how many seconds does the ball reach its maximum height? What is the maximum height? The maximum height occurs at the vertex. The ball reaches its maximum height, 271 ft, after 3.5 seconds. Verify with a graphing calculator.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 5(c) Solving a Problem Involving Projectile Motion (page 308) For what interval of time is the height of the ball greater than 200 ft? Solve the quadratic inequality. Use the quadratic formula to find the values of x that satisfy. a = –16, b = 112, c = –125 or

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 5(c) Solving a Problem Involving Projectile Motion (cont.) The two numbers divide a number line into three regions, Choose test values to see which interval satisfies the inequality. The ball will be greater than 200 ft above ground level between 1.39 and 5.61 seconds after it is thrown.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 5(c) Solving a Problem Involving Projectile Motion (cont.) Verify with a graphing calculator.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 5(d) Solving a Problem Involving Projectile Motion (page 308) After how many seconds will the ball hit the ground? Use the quadratic formula to find the positive solution of Reject The ball hits the ground after about 7.62 sec. Verify with a graphing calculator.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Synthetic Division 3.2 Synthetic Division ▪ Evaluating Polynomial Functions Using the Remainder Theorem ▪ Testing Potential Zeros

Properties for Using Synthetic Division Division Algorithm: Let f(x) and g(x) be polynomials with g(x) of lower degree than f(x). There exists a unique quotient, q(x) and remainder, r(x) where Special Case: For this case you can use synthetic division. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-20

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 1 Using Synthetic Division (page 323) Use synthetic division to divide. x – 3 is in the form x – k. Bring down the 4, and multiply 3(4) = 12.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 1 Using Synthetic Division (cont.) Add –15 and 12 to obtain –3. Multiply (–3)(3) = –9. Add 11 and –9 to obtain 2. Multiply (2)(3) = 6.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 1 Using Synthetic Division (cont.) Add –10 and 6 to obtain 4. Since the divisor x – k has degree 1, the degree of the quotient will always be one less than the degree of the original polynomial.

The Remainder Theorem If the polynomial f(x) is divided by x-k, then the remainder is equal to f(k). Zeros, Roots and x-intercepts: A zero of a polynomial function f is a number such that f(k) = 0 (the remainder). The real number zeros are the x-intercepts of the graph of the function. The Remainder Thm is a quick way to decide if a number k is a zero of f(x). A zero of f(x) is a root or solution of the equation f(x)=0. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-24

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 2 Applying the Remainder Theorem (page 324) Let. Use the remainder theorem to find f(4). Use synthetic division with k = 4. f(4) = 17 Insert 0 as the coefficient for the missing x 2 - term.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 3(a) Deciding Whether a Number is a Zero (page 325) Let. Is k = 2 a zero? Use synthetic division with k = 2. 2 is not a zero. Insert 0 as the coefficient for the missing x-term.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 3(b) Deciding Whether a Number is a Zero (page 325) Let. Is k = –3 a zero? Use synthetic division with k = –3. –3 is a zero.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 3(c) Deciding Whether a Number is a Zero (page 325) Let. Is k = 1 + 3i a zero? 1 + 3i is a zero. Use synthetic division with k = 1 + 3i.