Copyright © 2010 Pearson Education, Inc Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 1

Quadratic Functions and Equations 11 Quadratic Functions and Equations 11.1 Quadratic Equations 11.2 The Quadratic Formula 11.3 Studying Solutions of Quadratic Equations 11.4 Applications Involving Quadratic Equations 11.5 Equations Reducible to Quadratic 11.6 Quadratic Functions and Their Graphs 11.7 More About Graphing Quadratic Functions 11.8 Problem Solving and Quadratic Functions 11.9 Polynomial and Rational Inequalities Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Principle of Square Roots Completing the Square Problem Solving Quadratic Equations 11.1 The Principle of Square Roots Completing the Square Problem Solving Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Principle of Square Roots Let’s consider x2 = 25. We know that the number 25 has two real-number square roots, 5 and −5, the solutions of the equation. Thus we see that square roots can provide quick solutions for equations of the type x2 = k. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Principle of Square Roots For any real number k, if x2 = k, then Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve 5x 2 = 15. Give exact solutions and approximations to three decimal places. Solution Isolating x2 Using the principle of square roots We often use the symbol to represent both solutions. The solutions are which round to 1.732 and –1.732. The check is left to the student. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve 16x2 + 9 = 0. Recall that The solutions are The check is left to the student. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Principle of Square Roots (Generalized Form) For any real number k and any algebraic expression X: If X 2 = k, then Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve: (x + 3)2 = 7. The solutions are Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Completing the Square Not all quadratic equations can be solved as in the previous examples. By using a method called completing the square, we can use the principle of square roots to solve any quadratic equation. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution x2 + 10x + 25 = –4 + 25 Solve x2 + 10x + 4 = 0. Adding 25 to both sides. x2 + 10x + 25 = –4 + 25 Factoring Using the principle of square roots The solutions are Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

To Solve a Quadratic Equation in x by Completing the Square 1. Isolate the terms with variables on one side of the equation, and arrange them in descending order. 2. Divide both sides by the coefficient of x2 if that coefficient is not 1. 3. Complete the square by taking half of the coefficient of x and adding its square to both sides. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

6. Solve for x by adding or subtracting on both sides. 4. Express the trinomial as the square of a binomial (factor the trinomial) and simplify the other side. 5. Use the principle of square roots (find the square roots of both sides). 6. Solve for x by adding or subtracting on both sides. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Problem Solving After one year, an amount of money P, invested at 4% per year, is worth 104% of P, or P(1.04). If that amount continues to earn 4% interest per year, after the second year the investment will be worth 104% of P(1.04), or P(1.04)2. This is called compounding interest since after the first period, interest is earned on both the initial investment and the interest from the first period. Generalizing, we have the following. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Compound Interest Formula If the amount of money P is invested at interest rate r, compounded annually, then in t years, it will grow to the amount A given by (r is written in decimal notation.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution 6765 = 5800(1 + r)2 Jackson invested $5800 at an interest rate of r, compounded annually. In 2 yr, it grew to $6765. What was the interest rate? Solution Familiarize. We are already familiar with the compound-interest formula. 2. Translate. The translation consists of substituting into the formula: 6765 = 5800(1 + r)2 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

5. State. The interest rate was 8%. 3. Carry out. Solve for r: 6765/5800 = (1 + r)2 4. Check. Since the interest rate cannot negative, we need only to check .08 or 8%. If $5800 were invested at 8% compounded annually, then in 2 yr it would grow to 5800(1.08)2, or $6765. The number 8% checks. 5. State. The interest rate was 8%. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Using the Quadratic Formula Approximating Solutions 11.2 Solving Using the Quadratic Formula Approximating Solutions Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Using the Quadratic Formula Each time we solve by completing the square, the procedure is the same. When a procedure is repeated many times, a formula can often be developed to speed up our work. If we begin with a quadratic equation in standard form, ax2 + bx + c = 0, and solve by completing the square we arrive at the quadratic formula. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Quadratic Formula The solutions of ax2 + bx + c = 0, are given by Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve 3x2 + 5x = 2 using the quadratic formula. First determine a, b, and c: 3x2 + 5x – 2 = 0; a = 3, b = 5, and c = –2. Substituting Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The solutions are 1/3 and –2. The check is left to the student. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

To Solve a Quadratic Equation 1. If the equation can easily be written in the form ax2 = p or (x + k)2 = d, use the principle of square roots. 2. If step (1) does not apply, write the equation in the form ax2 + bx + c = 0. 3. Try factoring using the principle of zero products. 4. If factoring seems difficult or impossible, use the quadratic formula. Completing the square can also be used. The solutions of a quadratic equation can always be found using the quadratic formula. They cannot always be found by factoring. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Recall that a second-degree polynomial in one variable is said to be quadratic. Similarly, a second-degree polynomial function in one variable is said to be a quadratic function. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve x2 + 7 = 2x using the quadratic formula. First determine a, b, and c: x2 – 2x + 7 = 0; a = 1, b = –2, and c = 7. Substituting Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The check is left to the student. The solutions are The check is left to the student. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Approximating Solutions When the solution of an equation is irrational, a rational-number approximation is often useful. This is often the case in real-world applications similar to those found in section 8.4. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Use a calculator to approximate Solution Take the time to familiarize yourself with your calculator: Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Studying Solutions of Quadratic Equations 11.3 The Discriminant Writing Equations from Solutions Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Discriminant It is sometimes enough to know what type of number a solution will be, without actually solving the equation. From the quadratic formula, b2 – 4ac, is known as the discriminant. The discriminant determines what type of number the solutions of a quadratic equation are. The cases are summarized on the next slide. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Discriminant b2 – 4ac Nature of Solutions One solution; a rational number Positive Perfect square Not a perfect square Two different real-number solutions Solutions are rational Solutions are irrational conjugates Negative Two different imaginary-number solutions (complex conjugates) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution b2 – 4ac = (–1)2 – 4(4)(1) = –15. For the equation 4x2 – x + 1 = 0, determine what type of number the solutions are and how many solutions exist. Solution First determine a, b, and c: a = 4, b = –1, and c = 1. Compute the discriminant: b2 – 4ac = (–1)2 – 4(4)(1) = –15. Since the discriminant is negative, there are two imaginary-number solutions that are complex conjugates of each other. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution b2 – 4ac = (–10)2 – 4(5)(5) = 0. For the equation 5x2 – 10x + 5 = 0, determine what type of number the solutions are and how many solutions exist. Solution First determine a, b, and c: a = 5, b = –10, and c = 5. Compute the discriminant: b2 – 4ac = (–10)2 – 4(5)(5) = 0. There is exactly one solution, and it is rational. This indicates that 5x2 – 10x + 5 = 0 can be solved by factoring. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution b2 – 4ac = (7)2 – 4(2)(–3) = 73. For the equation 2x2 + 7x – 3 = 0, determine what type of number the solutions are and how many solutions exist. Solution First determine a, b, and c: a = 2, b = 7, and c = –3. Compute the discriminant: b2 – 4ac = (7)2 – 4(2)(–3) = 73. The discriminant is a positive number that is not a perfect square. Thus there are two irrational solutions that are conjugates of each other. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Writing Equations from Solutions We know by the principle of zero products that (x – 1)(x + 4) = 0 has solutions 1 and -4. If we know the solutions of an equation, we can write an equation, using the principle in reverse. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Find an equation for which 5 and –4/3 are solutions. x = 5 or x = –4/3 x – 5 = 0 or x + 4/3 = 0 Get 0’s on one side Using the principle of zero products (x – 5)(x + 4/3) = 0 x2 – 5x + 4/3x – 20/3 = 0 Multiplying 3x2 – 11x – 20 = 0 Combining like terms and clearing fractions Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Find an equation for which 3i and –3i are solutions. x = 3i or x = –3i x – 3i = 0 or x + 3i = 0 Get 0’s on one side Using the principle of zero products (x – 3i)(x + 3i) = 0 x2 – 3ix + 3ix – 9i2 = 0 Multiplying x2 + 9 = 0 Combining like terms Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Applications Involving Quadratic Equations 11.4 Solving Problems Solving Formulas Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Problems As we found in Section 6.7, some problems translate to rational equations. The solution of such rational equations can involve quadratic equations. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Cade traveled 48 miles at a certain speed. If he had gone 4 mph faster, the trip would have taken 1 hr less. Find Cade’s average speed. Solution Familiarize. As in Section 6.5, we can create a table. Let r represent the rate, in miles per hour, and t the time, in hours for Cade’s trip. Distance Speed Time 48 r t r + 4 t – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

and 2. Translate. From the table we obtain 3. Carry out. A system of equations has been formed. We substitute for r from the first equation into the second and solve the resulting equation: Multiplying by the LCD Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2010 Pearson Education, Inc Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

This is 1 hr less than the trip actually took, so the answer checks. 4. Check. Note that we solved for t, not r as required. Since negative time has no meaning here, we disregard the –3 and use 4 to find r: 12 mph. To see if 12 mph checks, we increase the speed 4 mph to 16 and see how long the trip would have taken at that speed: This is 1 hr less than the trip actually took, so the answer checks. 5. State. Cade traveled at 12 mph. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Formulas Recall that to solve a formula for a certain letter, we use the principles for solving equations to get that letter alone on one side. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve Solution Multiplying both sides by 2 Writing standard form Using the quadratic formula Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

To Solve a Formula for a Letter – Say, h 1. Clear fractions and use the principle of powers, as needed. Perform these steps until radicals containing h are gone and h is not in any denominator. 2. Combine all like terms. 3. If the only power of h is h1, the equation can be solved as in Sections 2.3 and 7.5. 4. If h2 appears but h does not, solve for h2 and use the principle of square roots to solve for h. 5. If there are terms containing both h and h2, put the equation in standard form and use the quadratic formula. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Equations Reducible to Quadratic 11.5 Recognizing Equations in Quadratic Form Radical Equations and Rational Equations Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Recognizing Equations in Quadratic Form Certain equations that are not really quadratic can be thought of in such a way that they can be solved as quadratic. For example, because the square of x2 is x4, the equation x4 – 5x2 + 4 = 0 is said to be “quadratic in x2”: Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

x4 – 5x2 + 4 = 0 (x2)2 – 5(x2) + 4 = 0 u2 – 5u + 4 = 0. The last equation can be solved by factoring or by the quadratic formula. Then, remembering that u = x2, we can solve for x. Equations that can be solved like this are reducible to quadratic, or in quadratic form. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve x4 – 5x2 + 4 = 0. Solution u2 – 5u + 4 = 0 Let u = x2. Then we solve by substituting u for x2 and u2 for x4: u2 – 5u + 4 = 0 (u – 1)(u – 4) = 0 Factoring Principle of zero products u – 1 = 0 or u – 4 = 0 u = 1 or u = 4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

x2 = 1 or x2 = 4 Replace u with x2 To check, note that for both x = 1 and x = –1, we have x2 = 1 and x4 = 1. Similarly, for both x = 2 and x = –2, we have x2 = 4 and x4 = 16. Thus instead of making four checks, we need make only two. Check: x = 1: x = 2: x4 – 5x2 + 4 = 0 x4 – 5x2 + 4 = 0 (16) – 5(4) + 4 = 0 (1) – 5(1) + 4 = 0 TRUE TRUE The solutions are 1, –1, 2, and –2. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Caution! A common error on problems like the previous example is to solve for u but forget to solve for x. Remember to solve for the original variable! Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Radical and Rational Equations Sometimes rational equations, radical equations, or equations containing exponents that are fractions are reducible to quadratic. It is especially important that answers to these equations be checked in the original equation. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve Solution u2 – 8u – 9 = 0 (u – 9)(u +1) = 0 Let u = . Then we solve by substituting u for and u2 for x: u2 – 8u – 9 = 0 (u – 9)(u +1) = 0 u – 9 = 0 or u + 1 = 0 u = 9 or u = –1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Check: x = 81: x = 1: The solution is 81. FALSE TRUE Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve Solution u2 + 4u – 2 = 0 Let u = t −1. Then we solve by substituting u for t −1 and u2 for t −2: u2 + 4u – 2 = 0 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The check is left to the student. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

To Solve an Equation That is Reducible to Quadratic 1. Look for two variable expressions in the equation. One expressions should be the square of the other. 2. Write down any substitutions that you are making. 3. Remember to solve for the variable that is used in the original equation. 4. Check possible answers in the original equation. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Quadratic Functions and Their Graphs 11.6 The Graph of f (x) = ax2 The Graph of f (x) = a(x – h)2 The Graph of f (x) = a(x – h)2 + k Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Graph of f (x) = ax2 All quadratic functions have graphs similar to y = x2. Such curves are called parabolas. They are U-shaped and symmetric with respect to a vertical line known as the parabola’s axis of symmetry. For the graph of f (x) = x2, the y-axis is the axis of symmetry. The point (0, 0) is known as the vertex of this parabola. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Graph: Solution 1 –1 2 –2 8 (0, 0) (1, 2) (–1, 2) (2, 8) (−2, 8) (2, 8) x f(x) = x2 (x, f(x)) 1 –1 2 –2 8 (0, 0) (1, 2) (–1, 2) (2, 8) (–2, 8) (−1, 2) (1, 2) (0,0) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Graph: Solution 1 –1 2 –2 –3 –12 (0, 0) (1, –3) (–1, –3) f(x) (x, f(x)) 1 –1 2 –2 –3 –12 (0, 0) (1, –3) (–1, –3) (2, –12) (–2, –12) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

4 3 6 2 5 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Graphing f (x) = ax2 The graph of f (x) = ax2 is a parabola with x = 0 as its axis of symmetry. Its vertex is the origin. For a > 0, the parabola opens upward. For a < 0, the parabola opens downward. If |a| is greater than 1, the parabola is narrower than y = x2. If |a| is between 0 and 1, the parabola is wider than y = x2. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Graph of f (x) = a(x – h)2 We could next consider graphs of f (x) = ax2 + bx + c, where b and c are not both 0. It turns out to be convenient to first graph f (x) = a(x – h)2, where h is some constant. This allows us to observe similarities to the graphs drawn in previous slides. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Graph: Solution vertex 1 –1 2 3 4 9 x f(x) = (x – 2)2 1 –1 2 3 4 9 vertex Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Graphing f (x) = a(x – h)2 The graph of f (x) = a(x – h)2 has the same shape as the graph of y = ax2. If h is positive, the graph of y = ax2 is shifted h units to the right. If h is negative, the graph of y = ax2 is shifted |h| units to the left. The vertex is (h, 0) and the axis of symmetry is x = h. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Graph of f (x) = a(x – h)2 + k Given a graph of f (x) = a(x – h)2, what happens if we add a constant k? Suppose that we add 2. This increases f (x) by 2, so the curve is moved up. If k is negative, the curve is moved down. The axis of symmetry for the parabola remains x = h, but the vertex will be at (h, k), or equivalently (h, f (h)). Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Graphing f (x) = a(x – h)2 + k The graph of f (x) = a(x – h)2 + k has the same shape as the graph of y = a(x – h)2. If k is positive, the graph of y = a(x – h)2 is shifted k units up. If k is negative, the graph of y = a(x – h)2 is shifted |k| units down. The vertex is (h, k), and the axis of symmetry is x = h. For a > 0, k is the minimum function value. For a < 0, the maximum function value is k. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Graph: and find the maximum function value. Solution –1 –2 –3 –4 –5 -11/2 –3/2 Maximum = −1 vertex Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

More About Graphing Quadratic Functions 11.7 Completing the Square Finding Intercepts Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Completing the Square By completing the square, we can rewrite any polynomial ax2 + bx + c in the form a(x – h)2 + k. Once that has been done, the procedures discussed in Section 11.6 will enable us to graph any quadratic function. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Graph: f (x) = x2 – 2x – 1 = (x2 – 2x) – 1 y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 = (x2 – 2x) – 1 = (x2 – 2x + 1 – 1) – 1 = (x2 – 2x + 1) – 1 – 1 = (x – 1)2 – 2 The vertex is at (1, –2). Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Graph: f (x) = –2x2 + 6x – 3 = –2(x2 – 3x) – 3 y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 = –2(x2 – 3x) – 3 = –2(x2 – 3x + 9/4 – 9/4) – 3 = –2(x2 – 3x + 9/4) – 3 + 18/4 = –2(x – 3/2)2 + 3/2 The vertex is at (3/2, 3/2). Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Vertex of a Parabola The vertex of a parabola given by f (x) = ax2 + bx + c is The x-coordinate of the vertex is –b/(2a). The axis of symmetry is x = -b/(2a). The second coordinate of the vertex is most commonly found by computing Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Finding Intercepts For any function f, the y-intercept occurs at f (0). Thus for f (x) = ax2 + bx + c, the y-intercept is simply (0, c). To find x-intercepts, we look for points where y = 0 or f (x) = 0. Thus, for f (x) = ax2 + bx + c, the x-intercepts occur at those x-values for which ax2 + bx + c = 0 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

f (x) = ax2 + bx + c y - intercept x - intercepts y x 6 5 4 3 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 -1 -2 x -3 -4 -5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Find the x- and y-intercepts of the graph of f (x) = x2 + 3x – 1. Solution The y-intercept is simply (0, f (0)), or (0, –1). To find the x-intercepts, we solve the equation: x2 + 3x – 1 = 0. Since we are unable to solve by factoring, we use the quadratic formula to get If graphing, we would approximate to get (–3.3, 0) and (0.3, 0). Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Problem Solving and Quadratic Functions 11.8 Maximum and Minimum Problems Fitting Quadratic Functions to Data Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Maximum and Minimum Problems We have seen that for any quadratic function f, the value of f (x) at the vertex is either a maximum or a minimum. Thus problems in which a quantity must be maximized or minimized can be solved by finding the coordinates of the vertex, assuming the problem can be modeled with a quadratic function. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example A farmer has 200 ft of fence with which to form a rectangular pen on his farm. If an existing fence forms one side of the rectangle, what dimensions will maximize the size of the area? Solution Familiarize. We make a drawing and label it, letting w = the width of the rectangle, in feet and l = the length of the rectangle, in feet. Existing fence w l Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Recall that Area = lw and Perimeter = 2w + 2l Recall that Area = lw and Perimeter = 2w + 2l. Since the existing fence forms one length of the rectangle, the fence will comprise three sides. Thus 2w + l = 200. 2. Translate. We have two equations: One guarantees that all 200 ft of fence will be used; the other expresses area in terms of length and width. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Factoring and completing the square, we get 3. Carryout. We need to express A as a function of l or w but not both. To do so, we solve for l in the first equation to obtain l = 200 – 2w. Substituting for l in the second equation, we get a quadratic function: A = (200 – 2w)w = 200w – 2w2 Factoring and completing the square, we get A = –2(w – 50)2 + 5000. The maximum area, 5000 ft2, occurs when w = 50 ft and l = 200 – 2(50), or 100 ft. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

4. Check. The check is left to the student. 5. State. The dimensions for the largest rectangular area for the pen that can be enclosed is 50 ft by 100 ft. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Fitting Quadratic Functions to Data Whenever a certain quadratic function fits a situation, that function can be determined if three inputs and their outputs are known. Each of the given ordered pairs is called a data point. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Use the data points (0, 10.4), (3, 16.8), and (6, 12.6) to find a quadratic function that fits the data. Solution We are looking for a function of the form f (x) = ax2 + bx + c, given that f (0) = 10.4, f (3) = 16.8, and f (6) = 12.6. Thus a(0)2 + b(0) + c = 10.4 a(3)2 + b(3) + c = 16.8 a(6)2 + b(6) + c = 12.6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Simplifying we see that we need to solve the system c = 10.4 (1) 9a + 3b + c = 16.8 (2) 36a + 6b + c = 12.6 (3) Substituting c = 10.4 into equations (2) and (3) and solving the resulting system we get So we have Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Polynomial and Rational Inequalities 11.9 Quadratic and Other Polynomial Inequalities Rational Inequalities Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Quadratic and Other Polynomial Inequalities Inequalities like the following are called polynomial inequalities: Second-degree polynomial inequalities in one variable are called quadratic inequalities. To solve polynomial inequalities, we often focus attention on where the outputs of a polynomial function are positive and where they are negative. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve: Solution Consider the related function f (x) = x2 + 2x – 8 and its graph. Since the graph opens upward, the y –values are positive outside the interval formed by the x-intercepts. x y -5 -4 -3 -2 -1 1 2 3 4 5 -6 4 -4 6 -2 2 8 -8 -10 -14 -12 y = x2 + 2x – 8 positive y-values Thus the solution set of the inequality is Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

In the next example, we simplify the process by concentrating on the sign of a polynomial function over each interval formed by the x-intercepts. We will do this by tracking the sign of each factor. By looking at how many positive or negative factors are being multiplied, we will be able to determine the sign of the polynomial function. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution We first solve the related equation: f (x) = 0 x3 – 9x = 0 x(x – 3)(x + 3) = 0 Factor Using the principle of zero products x = 0 or x – 3 = 0 or x + 3 = 0 x = 0 or x = 3 or x = –3. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The function f has zeros at –3, 0 and 3 The function f has zeros at –3, 0 and 3. We will use the factorization f (x) = x(x – 3)(x + 3). The product x(x – 3)(x + 3) is positive or negative, depending on the signs of x, x – 3, and x + 3. This is easily determined using a chart. Sign of x: Sign of x – 3: Sign of x + 3: Sign of product x(x – 3)(x + 3) –3 0 3 Interval: + – Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

A product is negative when it has an odd number of negative factors A product is negative when it has an odd number of negative factors. Since the sign allows for equality, the endpoints –3, 0, and 3 are solutions. From the chart, we see that the solution set is Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

To Solve a Polynomial Inequality Using Factors 1. Add or subtract to get 0 on one side and solve the related polynomial equation by factoring. 2. Use the numbers found in step (1) to divide the number line into intervals. 3. Using a test value from each interval, determine the sign of each factor over that interval. First find the sign of each factor, and then determine the sign of the product of the factors. Remember that the product of an odd number of negative numbers is negative. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

To Solve a Polynomial Inequality Using Factors 4. Select the interval(s) for which the inequality is satisfied and write set-builder notation or interval notation for the solution set. Include the endpoints of the intervals when is used. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Rational Inequalities Inequalities involving rational expressions are called rational inequalities. Like polynomial inequalities, rational inequalities can be solved using test values. Unlike polynomials, however, rational expressions often have values for which the expression is undefined. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve: Solution We write the related equation by changing the symbol to = : Next we solve the related equation: Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

In the case of rational inequalities, we must always find any values that make the denominator 0. As noted at the beginning that occurs when x = 3. Now we use 3 and 7 to divide the number line into intervals: 3 7 A B C We test a number in each interval to see where the original inequality is satisfied: Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

0 is not a solution, so interval A is not part of the solution set. A: Test 0, 4 is a solution, so interval B is part of the solution set. B: Test 4, 10 is not a solution, so interval C is not part of the solution set. C: Test 10, Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The solution set includes the interval B The solution set includes the interval B. The endpoint 7 is included because the inequality symbol is and 7 is a solution of the related equation. The number 3 is not included because (x + 5)/(x – 3) is undefined for x = 3. This the solution set of the original inequality is Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

To Solve a Rational Inequality 1.. Find any replacements for which the rational expression is undefined. 2. Change the inequality symbol to an equals sign and solve the related equation 3. Use the numbers found in step (1) and (2) to divide the number line into intervals. 4. Substitute a test value from each interval into the inequality. If the number is a solution, then the interval to which it belongs is part of the solution set. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

To Solve a Rational Inequality 5. Select the interval(s) and any endpoints for which the inequality is satisfied and write set-builder notation or interval notation for the solution set. If the inequality symbol is then the solutions from step (2) are also included in the solution set. Those numbers found in step (1) should be excluded from the solution set, even if they are solutions from step (2). Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley