Slide 8- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Quadratic Functions and Equations 11.1Quadratic Equations 11.2The Quadratic Formula 11.3Studying Solutions of Quadratic Equations 11.4Applications Involving Quadratic Equations 11.5Equations Reducible to Quadratic 11.6Quadratic Functions and Their Graphs 11.7More About Graphing Quadratic Functions 11.8Problem Solving and Quadratic Functions 11.9Polynomial and Rational Inequalities 11

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

Slide 8- 4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Principle of Square Roots Let’s consider x 2 = 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 x 2 = k.

Slide 8- 5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Principle of Square Roots For any real number k, if x 2 = k, then

Slide 8- 6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve 5x 2 = 15. Give exact solutions and approximations to three decimal places. We often use the symbol to represent both solutions. The solutions are which round to and – The check is left to the student. Isolating x 2 Using the principle of square roots Example

Slide 8- 7 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve 16x = 0. The solutions are The check is left to the student. Recall that Example

Slide 8- 8 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

Slide 8- 9 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve: (x + 3) 2 = 7. The solutions are Example

Slide 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.

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

Slide 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 x 2 if that coefficient is not Complete the square by taking half of the coefficient of x and adding its square to both sides.

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

Slide 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.

Slide 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.)

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

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

Slide 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, ax 2 + bx + c = 0, and solve by completing the square we arrive at the quadratic formula.

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Quadratic Formula The solutions of ax 2 + bx + c = 0, are given by

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

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

Slide 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 ax 2 = 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 ax 2 + 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.

Slide 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.

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

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

Slide 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.

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

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

Slide 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, b 2 – 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.

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Discriminant b 2 – 4ac Nature of Solutions 0One solution; a rational number Positive Perfect square Not a perfect square Two different real-number solutions Solutions are rational Solutions are irrational conjugates NegativeTwo different imaginary-number solutions (complex conjugates)

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution For the equation 4x 2 – x + 1 = 0, determine what type of number the solutions are and how many solutions exist. First determine a, b, and c: a = 4, b = –1, and c = 1. Compute the discriminant: b 2 – 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. Example

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

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution For the equation 2x 2 + 7x – 3 = 0, determine what type of number the solutions are and how many solutions exist. First determine a, b, and c: a = 2, b = 7, and c = –3. Compute the discriminant: b 2 – 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. Example

Slide 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.

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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 (x – 5)(x + 4/3) = 0 x 2 – 5x + 4/3x – 20/3 = 0 3x 2 – 11x – 20 = 0 Get 0’s on one side Using the principle of zero products Multiplying Combining like terms and clearing fractions Example

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find an equation for which 3i and –3i are solutions. x = 3i or x = –3i x – 3i = 0 or x + 3i = 0 (x – 3i)(x + 3i) = 0 x 2 – 3ix + 3ix – 9i 2 = 0 x = 0 Get 0’s on one side Using the principle of zero products Multiplying Combining like terms Example

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

Slide 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.

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution 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. 1.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. DistanceSpeedTime 48rt r + 4t – 1 Example

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. From the table we obtain and 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

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

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

Slide 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.

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

Slide 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 h 1, the equation can be solved as in Sections 2.3 and If h 2 appears but h does not, solve for h 2 and use the principle of square roots to solve for h. 5.If there are terms containing both h and h 2, 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 Recognizing Equations in Quadratic Form Radical Equations and Rational Equations 11.5

Slide 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 x 2 is x 4, the equation x 4 – 5x = 0 is said to be “quadratic in x 2 ”:

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

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve x 4 – 5x = 0. u 2 – 5u + 4 = 0 Let u = x 2. Then we solve by substituting u for x 2 and u 2 for x 4 : (u – 1)(u – 4) = 0 u = 1 or u = 4 u – 1 = 0 or u – 4 = 0 Factoring Principle of zero products Example

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

Slide 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!

Slide 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.

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve u 2 – 8u – 9 = 0 (u – 9)(u +1) = 0 u = 9 or u = –1 u – 9 = 0 or u + 1 = 0 Let u =. Then we solve by substituting u for and u 2 for x: Example

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

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

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

Slide 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 The Graph of f (x) = ax 2 The Graph of f (x) = a(x – h) 2 The Graph of f (x) = a(x – h) 2 + k 11.6

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Graph of f (x) = ax 2 All quadratic functions have graphs similar to y = x 2. 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) = x 2, the y-axis is the axis of symmetry. The point (0, 0) is known as the vertex of this parabola.

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

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

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

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing f (x) = ax 2 The graph of f (x) = ax 2 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 = x 2. If |a| is between 0 and 1, the parabola is wider than y = x 2.

Slide 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) = ax 2 + 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.

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph: Solution xf(x) = (x – 2) – vertex Example

Slide 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 = ax 2. If h is positive, the graph of y = ax 2 is shifted h units to the right. If h is negative, the graph of y = ax 2 is shifted |h| units to the left. The vertex is (h, 0) and the axis of symmetry is x = h.

Slide 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)).

Slide 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.

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Graph: and find the maximum function value. x 0 –1 –2 –3 –4 –5 -11/2 –3 –3/2 –1 –3/2 –3 vertex Example Maximum = −1

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

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Completing the Square By completing the square, we can rewrite any polynomial ax 2 + 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.

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

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

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Vertex of a Parabola The vertex of a parabola given by f (x) = ax 2 + 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

Slide 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) = ax 2 + 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) = ax 2 + bx + c, the x- intercepts occur at those x-values for which ax 2 + bx + c = 0

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley x y f (x) = ax 2 + bx + c x - intercepts y - intercept

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find the x- and y-intercepts of the graph of f (x) = x 2 + 3x – 1. The y-intercept is simply (0, f (0)), or (0, –1). To find the x-intercepts, we solve the equation: x 2 + 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). Example

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

Slide 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.

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

Slide 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.

Slide Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use the data points (0, 10.4), (3, 16.8), and (6, 12.6) to find a quadratic function that fits the data. Example