pg098 V G2 5 36058 HCG Cannon Elich cr 10 31 95 QC. 98 Chapter 2 Functions,I had mathematical Definition quadratic function. curiosity very early My A quadratic function is a function with an equation equivalent to. father had in his library a,wonderful series of f x 5 ax 2 1 bx 1 c 1. German paperback, books One was Euler s where a b and c are real numbers and a is not zero. Algebra I discovered by, myself how to solve For example g x 5 5 2 4x 2 and h x 5 2x 2 1 2 2 x 2 are quadratic func. equations I remember that tions but F x 5 x 2 1 x 1 1 and G x 5 x 2 3 2 2 x 2 are not In fact G is. I did this by an incredible,concentration and almost. painful and not quite, Basic Transformations and Graphs of Quadratic Functions. conscious effort What I did, amounted to completing The graphing techniques we introduced in Section 2 3 are collectively called basic. the square in my head,without paper or pencil, transformations The graph of any linear function is a line and we will show that. Stan Ulam that graph of any quadratic function can be obtained from the core parabola. f x 5 x 2 by applying basic transformations We apply terminology from the core. parabola to parabolas in general The point 0 0 is called the vertex of the core. parabola and the y axis is the axis of symmetry The axis of symmetry is a help. in making a hand sketch of a parabola Whenever we locate a point of the parabola. on one side of the axis of symmetry we automatically have another point located. symmetrically on the other side, We will derive a transformation form for a general quadratic function an. equation that identifies the vertex and axis of symmetry of the graph but to graph. any particular quadratic you may not need all of the steps Indeed because a. graphing calculator graphs any quadratic function we could ask why we need the. transformation form at all Because a calculator graph is dependent on the window. we choose and the pixel coordinates we need an algebraic form from which we can. read more exact information, To begin we factor out the coefficient a from the x terms and then add and. subtract the square of half the resulting x coefficient to complete the square on x. f x 5 ax 2 1 bx 1 c 5 a x 2 1 x 1 c,5 a x2 1 x 1 2 2 2 1 c. The final equation has the form,f x 5 a x 2 h 2 1 k 2. which we recognize as a core parabola shifted so that the vertex is at the point h k. and the axis of symmetry is the line x 5 h,Parabola Features. Looking at the derivation of Equation 2 we can make some observations about. the graphs of quadratic functions, pg99 R G1 5 36058 HCG Cannon Elich cr 11 14 95 QC4. 2 5 Quadratic Functions Parabolas and Problem Solving 99. Graphs of quadratic functions, For the quadratic function f x 5 ax 2 1 bx 1 c The graph is a parabola. Axis of with axis of symmetry x 5,The parabola opens upward if a 0 downward if a 0. To find the coordinates of the vertex set x 5,x 2a Then the y coordinate is. given by y 5 f,x intercept, The graph of every quadratic function intersects the y axis where x 5 0 but it. need not have any x intercept points To find any x intercepts we solve the equation. y intercept Vertex f x 5 0 By its nature every quadratic function has a maximum or a minimum. point depending on whether the parabola opens down or up that occurs at the vertex of. FIGURE 26 the parabola See Figure 26, cEXAMPLE 1 Locating the vertex of a parabola The graph of the quad. ratic function is a parabola Locate the vertex in two ways i by writing the. function in the form of Equation 2 and ii by setting x 5 2 Sketch the. 2 a f x 5 x 2 2 2x 2 3 b f x 5 2x 2 1 4x 2 1,1 y x2 Solution. 4 3 2 11 1 2 3 4 a i Complete the square on the x terms. 2 f x 5 x 2 2 2x 2 3 5 x 2 2 2x 1 1 2 1 2 3,5 x 2 1 2 2 4. a f x x 2 2x 3 From this form the graph is the core parabola shifted 1 unit right and 4. x 1 2 4 units down the vertex is at 1 24, y ii From f x 5 x 2 2 2x 2 3 2 2ab 5 2222 1 5 1 Substituting 1 for x. f 1 5 1 2 2 1 2 3 5 24 so again the vertex is the point 1 24 The. 4 graph is shown in Figure 27 a, 3 b i Before completing the square we factor out 2 from the x terms. 1 f x 5 2x 2 1 4x 2 1 5 2 x 2 1 2x 2 1,x 5 2 x 2 1 2x 1 1 2 1 2 1 5 2 x 1 1 2 2 3. The graph is obtained by shifting the core parabola 1 unit left stretching. by a factor of 2 and translating the stretched parabola down 3 units the. 1 3 vertex is at 21 23,b f x 2x 2 4x 1, ii From f x 5 2x 2 1 4x 2 1 2 2ab 5 2 44 5 21 and f 21 5 2 21 2 1. 2 x 1 2 3 4 21 2 1 5 23 and the vertex is at 21 23 The graph appears in. Figure 27 b b, cEXAMPLE 2 Graph maximum or minimum and intercept points. Find the maximum or minimum value of f and the intercept points both from a. calculator graph and algebraically,a f x 5 x 2 2 2x 2 3 b f x 5 2x 2 1 4x 2 1. pg100 V G2 5 36058 HCG Cannon Elich jn 11 22 95 MP1. 100 Chapter 2 Functions, a From Example 1 a and the graph in Figure 27 a the minimum value of f is. the y coordinate of the vertex 24 For the y intercept f 0 5 23 so the. y intercept point is 0 23 Solving the equation x 2 2 2x 2 3 5 0 we have. x 2 3 x 1 1 5 0 so the x intercept points are 3 0 and 21 0 While a. calculator graph in any window shows that the x intercept points are near. x 5 3 and x 5 21 in trace mode we do not see the x intercepts exactly. where the y coordinate equals 0 unless we have a decimal window We can. zoom in as needed though to get as close to 3 0 and 21 0 as desired. b From Example 1 b and the graph in Figure 27 b the function has a mini. mum value of 23 Since f 0 5 21 the y intercept point is 0 21 Tracing. along a calculator graph we find that the x intercepts are near 0 2 and 22 2. We can get closer approximations by zooming in but the equation. 2x 2 1 4x 2 1 5 0 does not factor with rational numbers so we cannot read. exact coordinates of the x intercept points on any calculator graph We use the. quadratic formula to solve the equation and find the x intercept points exactly. Strategy a Draw a sepa Quadratic Functions with Limited Domain. rate diagram Figure 28b to, show a right triangle formed According to the domain convention the domain of any quadratic function is the set. by altitude CD nCDB is of all real numbers unless there is some restriction Many applications place. similar to nFEB Use ratios natural restrictions on domains as illustrated in the next two examples. of the sides to relate h and x, cEXAMPLE 3 Limited domain A rectangle is inscribed in an isosceles tri. angle ABC as shown in Figure 28a where AB 5 6 and AC 5 BC 5 5 Let. x denote the width h the height and K the area of the rectangle Find an equation. for a h as a function of x b K as a function of x c Find the domain of each. D a Following the strategy the ratios and are equal DB 5 3. A 6 B DB EB, a FE 5 h and EB 5 3 2 For CD we can apply the Pythagorean. theorem to nCDB CD 5 52 2 32 5 4 Therefore,FE CD h 4 2x. 5 or 5 or h 5 4 2,F EB DB x 3 3,h b The area K is the product of x and h. D E B K 5 x h5 x 42 5 4x 2 x 2, c From the nature of the problem there is no rectangle unless x is a positive. FIGURE 28 number less than 6 Hence the domain of both h and K is x 0 x 6 b. pg101 R G1 5 36058 HCG Cannon Elich jn 11 22 95 MP1. 2 5 Quadratic Functions Parabolas and Problem Solving 101. cEXAMPLE 4 Minimizing area Find the dimensions x and h for the. rectangle with the maximum area that can be inscribed in the isosceles triangle. ABC in Figure 28a,Strategy Graph K and Solution, find the highest point on Follow the strategy In Example 3 we found the area K as a quadratic function of x. the graph the vertex,K x 5 2 x 2 1 4x 0 x 6, Graphing K as a function of x we get part of a parabola that opens down The graph. of K is shown in Figure 29 The maximum value of K occurs at the vertex of the. parabola where, When x 5 3 h 5 4 2 2x3 5 4 2 23 3 5 2 and K 3 5 6 Therefore the inscribed. rectangle with the largest area has sides of lengths 3 and 2 and area 6 b. Solving Quadratic Inequalities, In Section 1 5 we solved quadratic inequalities by factoring quadratic expressions. Here we look at the more general situation of solving quadratic inequalities with. the use of graphs as illustrated in the following example. K x 2 x2 4x,6 A 1 5 0 B 1 5 0,2 1 1 2 3 4 5 6,1 1 2 3 4 5 6 7. FIGURE 29 FIGURE 30, cEXAMPLE 5 Using a graph Find the domain of the function f x 5. x 2 2 2x 2 4,Strategy Use a graph of Solution, y 5 x 2 2 2x 2 4 to see Follow the strategy Let y 5 x 2 2 2x 2 4 and draw a graph This gives a parabola. where the y values are non that opens upward with vertex at 1 25 To find the x intercept points solve the. x 2 2 2x 2 4 5 0, by the quadratic formula to get x 5 1 6 5 Thus the intercept points are. A 1 2 5 0 and B 1 1 5 0 as shown in Figure 30 Use the graph to read off. the solution set to the inequality that defines the domain of f. D 5 x x 1 2 5 or x 1 1 5,5 2 1 2 5 1 1 5 b, pg102 V G2 5 36058 HCG Cannon Elich jb 11 9 95 QC2. 102 Chapter 2 Functions,Maximum and Minimum Values of a Function. When we use mathematical models to answer questions about an applied problem. we frequently need to determine the maximum or minimum values of a function. The general problem can be very difficult even using the tools of calculus but. when quadratic functions are involved one can simply read off a maximum or. minimum value from a graph,Definition maximum or minimum value of a function. Suppose f is a function with domain D, If there is a number k in D such that f k f x for every x in D then f k. is the maximum value of f, If there is a number k in D such that f k f x for every x in D then f k. is the minimum value of f, Strategy First draw a cEXAMPLE 6 A function with limited domain For the function f x 5. graph being aware of the x 2 2 4x 1 2 with domain D 5 x 0 x 52 find a the maximum and min. given domain From the imum values of f and b the set of values where f x 0. graph read off the minimum,or maximum values of y Solution. a The graph of f is the part of the parabola y 5 x 2 2 4x 1 2 on the interval. 0 52 A calculator graph may not allow us to read all needed points in exact. form so we first locate the vertex and then evaluate the function at the ends of. the domain,f x x2 4x 2, 5 Using x 5 2b 2a 5 2 5 2 we have f 2 5 22 so the point 2 22 is the. 0 2 vertex At the endpoints of the domain f 0 5 2 and f 52 5 2 74 so the graph. has endpoints 0 2 and 52 2 74, We get the graph shown in Figure 31 The maximum value of f is 2 which. 4 occurs at the left end of the graph and the minimum value is 22 at the vertex. 2 2 of the parabola, b To solve the inequality f x 0 we need the x intercept point between 0 and. FIGURE 31 1 From the quadratic formula f x 5 0 when x 5 2 6 2 Since the func. tion is only defined on the interval 0 52 f x 0 on the interval. cEXAMPLE 7 Ranges of transformed functions For the function f x 5. x 2 2 4x 1 2 with domain D 5 x 0 x 52 find the range of. a y 5 f x 1 2 b y 5 2 f x c y 5 f x, a The graph of y 5 f x 1 2 is shifted 2 units up from the graph in Figure 31. and is shown in Figure 32a The range of f is the interval 22 2 so the range. of y 5 f x 1 2 is also shifted 2 units up to 0 4, b To get the graph of y 5 2 f x we reflect the graph of f in the x axis as in. Figure 32b The range is still 22 2, c Stretching the graph of f vertically by a factor of 32 we get part of another. parabola y 5 32 x 2 2 6x 1 3 as shown in Figure 32c The range of y 5 32 f x. pg103 R G1 5 36058 HCG Cannon Elich jn 11 22 95 MP1. 2 5 Quadratic Functions Parabolas and Problem Solving 103. 52 14 x y 3,y f x 2 y f x 52 218,6 0 6 FIGURE 32,cEXAMPLE 8 Distance from a point to a line. 3 a Find the minimum distance from the origin to the line L given by 2x 1 y 5 6. 2 P u v b What are the coordinates of the point Q on L that is closest to the origin. 3 0 Solution, 1 1 2 3 4 5 First draw a diagram that will help formulate the problem Figure 33 Since. 0 0 P u v is on L then 2u 1 v 5 6 or v 5 6 2 2u, FIGURE 33 d 5 u 2 0 2 1 v 2 0 2 5 u 2 1 v 2 5 u 2 1 6 2 2u 2. z 5 5u 2 2 24u 1 36,z 5u2 24u 36, a The minimum value of d will occur when the expression under the radical is a. minimum Determine the minimum of the function,7 z 5 5u 2 2 24u 1 36. whose graph is shown in Figure 34 The lowest point on the parabola occurs. 1 1 2 3 4 u52 52 5 2 4, When u 5 2 4 z 5 7 2 Therefore the minimum value of z is 7 2 so the. 4 mi minimum distance from the origin to the line L is 7 2 2 68. Island b The point Q on L that is closest to the origin is given by u 5 2 4 and. x v 5 6 2 2 2 4 5 1 2 Thus Q is point 2 4 1 2 b,Looking Ahead to Calculus. Pipeline 15 mi Not all problems lead to quadratic functions The applied problem in the next. example requires calculus techniques to find an exact solution With a graphing. calculator however we can find an excellent approximation from a graph of a cost. cEXAMPLE 9 Reading a solution from a graph A freshwater pipeline is. to be built from a source on shore to an island 4 miles offshore as located in the. diagram Figure 35 The cost of running the pipeline along the shore is 7500 per. FIGURE 35 mile but construction offshore costs 13 500 per mile. pg104 V G2 5 36058 HCG Cannon Elich jb 11 8 95 QC1. 104 Chapter 2 Functions, a Express the construction cost C in thousands of dollars as a function of x. 2 5 Quadratic Functions Parabolas and Problem Solving 99 Graphs of quadratic functions For the quadratic functionf x 5 ax2 1 bx 1 c The graph is a parabola with axis of symmetry x 5 2b 2a The parabola opensupward if a 0 downward if a 0 To nd the coordinates of the vertex set x 5 2b 2a Thenthey coordinate is given by y 5 fS 2b 2a D The graph of every quadratic function intersects

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