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TH EDITION LIAL HORNSBY SCHNEIDER COLLEGE ALGEBRA

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Quadratic Equations and Models Quadratic Equations Graphing Techniques Completing the Square The Vertex Formula Quadratic Models and Curve Fitting

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Polynomial Function A polynomial function of degree n, where n is a nonnegative integer, is a function defined by an expression of the form where a n, a n-1, …, a 1, and a 0 are real numbers, with a n ≠ 0.

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Polynomial Function For the polynomial function defined by n is 3 and the polynomial has the form

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Quadratic Function A function is a quadratic function if where a, b, and c are real numbers, with a ≠ 0.

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Simplest Quadratic x (x)(x) – 2– 24 – 1– – 2– 2 3 – 2– 2 4 – 3– 3 – 4– 4 – 3– 3 – 4– 4 4 range [0, ) domain (− , ) x y

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Simplest Quadratic Parabolas are symmetric with respect to a line. The line of symmetry is called the axis of the parabola. The point where the axis intersects the parabola is the vertex of the parabola. Vertex Axis Opens up Opens down

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Applying Graphing Techniques to a Quadratic Function The graph of g (x) = ax 2 is a parabola with vertex at the origin that opens up if a is positive and down if a is negative. The width of the graph of g (x) is determined by the magnitude of a. The graph of g (x) is narrower than that of (x) = x 2 if a > 1 and is broader (wider) than that of (x) = x 2 if a < 1. By completing the square, any quadratic function can be written in the form the graph of F(x) is the same as the graph of g (x) = ax 2 translated h units horizontally (to the right if h is positive and to the left if h is negative) and translated k units vertically (up if k is positive and down if k is negative).

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Example 1 GRAPHING QUADRATIC FUNCTIONS Solution a. Graph the function. Give the domain and range. x (x)(x) – 1– 13 0– 2– 2 1– 5– 5 2– 6– 6 3– 5– 5 4– 2– – 2– 2 – 6– 6 Domain (− , ) Range [– 6, )

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Example 1 GRAPHING QUADRATIC FUNCTIONS Solution b. Graph the function. Give the domain and range. Domain (− , ) Range (– , 0] 2 3 – 2– 2 – 6– 6

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Example 1 GRAPHING QUADRATIC FUNCTIONS Solution c. Graph the function. Give the domain and range. Domain (− , ) Range (– , 3] (4, 3) 3 – 2– 2 – 6– 6 x = 4

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Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Express x 2 – 6x + 7 in the form (x– h) 2 + k by completing the square. Graph by completing the square and locating the vertex. Complete the square.

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Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Express x 2 – 6x + 7 in the form (x– h) 2 + k by completing the square. Graph by completing the square and locating the vertex. Add and subtract 9. Regroup terms. Factor; simplify. This form shows that the vertex is (3, – 2)

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Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Graph by completing the square and locating the vertex. Find additional ordered pairs that satisfy the equation. Use symmetry about the axis of the parabola to find other ordered pairs. Connect to obtain the graph. Domain is (− , )Range is [–2, )

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Motion Problems Note In Example 2 we added and subtracted 9 on the same side of the equation to complete the square in Section 1.4. This differs from adding the same number to each side of the equation, as when we completed the square. Since we want (x) (or y) alone on one side of the equation, we adjusted that step in the process of completing the square.

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution To complete the square, the coefficient of x 2 must be 1. Graph by completing the square and locating the vertex. Factor – 3 from the first two terms.

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Graph by completing the square and locating the vertex. Distributive property Be careful here. Factor; simplify.

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Graph by completing the square and locating the vertex. Factor; simplify.

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Intercepts are good additional points to find. Here is the y-intercept. Graph by completing the square and locating the vertex. Let x = 0.

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution The x-intercepts are found by setting (x) equal to 0 in the original equation. Graph by completing the square and locating the vertex. Let (x) = 0. Multiply by –1; rewrite. Factor. Zero-factor property

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE 2 2 The y-intercept is 1 This x-intercept is – 1. This x-intercept is 1/3.

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Graph of a Quadratic Function The quadratic function defined by (x) = ax 2 + bx + c can be written as where

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Graph of a Quadratic Function The graph of has the following characteristics. 1.It is a parabola with vertex (h, k) and the vertical line x = h as axis. 2.It opens up if a > 0 and down is a < 0. 3.It is broader than the graph of y = x 2 if a 1. 4.The y-intercept is (0) = c. 5.If b 2 – 4ac > 0, the x-intercepts are If b 2 – 4ac = 0, the x-intercepts is If b 2 – 4ac < 0, there are no x-intercepts.

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Example 4 FINDING THE AXIS AND THE VERTEX OF A PARABOLA USING THE VERTEX FORMULA Solution Here a = 2, b = 4, and c = 5. The axis of the parabola is the vertical line Find the axis and vertex of the parabola having equation (x) = 2x 2 +4x + 5 using the vertex formula. The vertex is (– 1, (– 1)). Since (– 1) = 2(– 1) (– 1) +5 = 3, the vertex is (– 1, 3).

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Quadratic Models and Curve Fitting Quadratic functions make good models for data sets where the data either increases, levels off, and then decreases, levels off, and then increases. An application that models the path of the projectile is

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution Use the projectile height function with v 0 = 80 and s 0 = 100. a. Give the function that describes the height of the ball in terms of time t. A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec.

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution One choice for a window is [-.3, 9.7] by [-60,300] b. Graph this function on a graphing calculator so that the y-intercept, the positive x- intercept, and the vertex are visible. A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec.

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution When 4.8 seconds have elapsed, the projectile is at a height of ft. c. The point (4.8, ) lies on the graph of the function. What does this mean for this particular situation? A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec.

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution Find the coordinates of the vertex of the parabola. a = – 16 and b = 80 d. After how many seconds does the projectile reach its maximum height? What is this maximum height? A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec.

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution Find the coordinates of the vertex of the parabola. a = – 16 and b = 80 d. After how many seconds does the projectile reach its maximum height? What is this maximum height? A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. and After 2.5 sec the ball reaches its maximum height of 200 ft.

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution Solve the quadratic inequality. e. For what interval of time is the height of the ball greater than 160 ft? A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. Subtract 160. Divide by – 4; reverse the inequality symbol.

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution By the quadratic formula, the solutions are… e. For what interval of time is the height of the ball greater than 160 ft? A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec.

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution The intervals are (– ,.92), (.92, 4.08), and (4.08, ). A test value in each interval shows that (.92, 4.08) satisfies the inequality. The ball is more than 160 ft above ground between.92 sec and 4.08 sec. e. For what interval of time is the height of the ball greater than 160 ft? A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec.

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution The height is zero when the ball hits the ground. Find the positive solution… f. After how many seconds will the ball hit the ground? A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec.

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Example 5 SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Solution f. After how many seconds will the ball hit the ground? A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. The ball hits the ground after about 6.04 sec.

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Example 6 MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS a. Prepare a scatter diagram, and determine a quadratic model for these data. 80 represents 1980, 100 represents 2000, and so on, and the number of outpatient visits is given in millions. YearVisitsYearVisits

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Example 6 MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS a. Prepare a scatter diagram, and determine a quadratic model for these data. 80 represents 1980, 100 represents 2000, and so on, and the number of outpatient visits is given in millions. Solution

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Example 6 MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS b. Predict the number of visits in represents 1980, 100 represents 2000, and so on, and the number of outpatient visits is given in millions. Solution

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Example 6 MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS b. Predict the number of visits in represents 1980, 100 represents 2000, and so on, and the number of outpatient visits is given in millions. Solution Since 2008 corresponds to x = 108, the model predicts that in 2008 the number of visits will be…

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Example 6 MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS b. Predict the number of visits in 2008.

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