Quadratic Function Graphs

12 Key excerpts on "Quadratic Function Graphs"

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  Intermediate Algebra

  • Mark D. Turner, Charles P. McKeague(Authors)
  • 2016(Publication Date)
  • XYZ Textbooks

    (Publisher)

  In this section, we will explore further these types of functions and their graphs. Up to now in this chapter, we have been studying quadratic equations. If we take a quadratic equation in standard form and use it instead as the formula for a function, the result is called a quadratic function . Here is a formal definition: Any function that can be written in the form f ( x ) = a x 2 + bx + c where a , b , and c are constants with a ≠ 0, is called a quadratic function . We refer to this form as standard form . quadratic function DEFINITION The Basic Quadratic Function The simplest of all quadratic functions results when we let a = 1 and b = c = 0. We will refer to f ( x ) = x 2 as the basic quadratic function . Table 1 gives some ordered pairs for this function, and the corresponding graph is shown in Figure 1. This graph is an example of a parabola . From the graph we can see that the domain of this function is the set of all real numbers and the range is { y | y ≥ 0}. The point (0, 0) where the parabola changes direction (the lowest point on the parabola) is called the vertex . FIGURE 1 –5 –4 –3 –2 –1 1 2 3 4 5 –5 –4 –3 –2 –1 1 2 3 4 5 Vertex y 5 x 2 2 3 -5 -4 -3 -2 -1 1 2 3 4 5 -1 1 2 3 5 6 7 8 9 x y Vertex f ( x ) = x 2 Axis of Symmetry Axis of Symmetry 4 x f ( x ) = x 2 − 3 9 − 2 4 − 1 1 0 0 1 1 2 4 3 9 TABLE 1 608 CHAPTER 8 Quadratic Equations and Functions We also observe that the graph is symmetric about the y -axis, meaning the right half of the graph is a mirror image of the left half. For this reason, the vertical line x = 0 passing through the vertex is called the axis of symmetry . All quadratic functions have a graph that is a parabola and a domain that is the set of all real numbers. However, the shape, direction, and position of the parabola can vary as we will see in the following segment. Transformations Let's consider quadratic functions of the form f ( x ) = a x 2 .

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  Algebra

  Form and Function

  • William G. McCallum, Eric Connally, Deborah Hughes-Hallett(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  Chapter Three QUADRATIC FUNCTIONS © Patrick Zephyr/Patrick Zephyr Nature Photography Contents 3.1 Introduction to Quadratic Functions. . . . . . . . . 100 Creating Computer Graphics . . . . . . . . 101 3.2 Quadratic Expressions . . . . . . . . . . . . . . . . . . . . 103 Interpreting Factored Form . . . . . . . . . . . . . . . . 104 Interpreting Vertex Form . . . . . . . . . . . . . . . . . . 106 Constructing Quadratic Functions . . . . . . . . . . . 107 3.3 Converting to Factored and Vertex Form . . . . . 111 Converting to Factored Form. . . . . . . . . . . . . . . 111 How Do We Put an Expression in Vertex Form? . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Visualizing Completing the Square . . . . . . . . . . 114 3.4 Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . 116 Solving Equations by Factoring . . . . . . . . . . . . . 116 Solving Equations with Perfect Squares . . . . . . 117 Solving by Completing the Square . . . . . . . . . . 119 The Quadratic Formula . . . . . . . . . . . . . . . . . . . 119 The Discriminant . . . . . . . . . . . . . . . . . . . . . . . . 121 3.5 Factoring Hidden Quadratics . . . . . . . . . . . . . . . 124 3.6 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . 129 Using Complex Numbers to Solve Equations . . 129 Algebra of Complex Numbers . . . . . . . . . . . . . . 130 Addition and Subtraction of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . 131 Multiplication of Complex Numbers . . 131 Division of Complex Numbers. . . . . . . 133 REVIEW PROBLEMS . . . . . . . . . . . . . . . . . . . . 134 SOLVING DRILL . . . . . . . . . . . . . . . . . . . . . . . . 138 100 Chapter 3 QUADRATIC FUNCTIONS 3.1 INTRODUCTION TO QUADRATIC FUNCTIONS The graph of a linear function is a straight line. If we want the graph to curve, we need a different sort of function. For example, Figure 3.1 shows the height of a ball thrown off the top of a building  seconds after it has been thrown.

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  Intermediate Algebra

  Concepts with Applications

  • Charles P. McKeague(Author)
  • 2013(Publication Date)
  • XYZ Textbooks

    (Publisher)

  The graph of the simplest quadratic function f (x) = x 2 appears as a parabola with its vertex at the origin (0, 0) because the value for a is 1 and the value for b is 0. (See Figure 2.) We can use the vertex point along with the x- and y-intercepts to sketch the graph of any equation of the form y = ax 2 + bx + c. FIGURE 2 –5 –4 –3 –2 –1 1 2 3 4 5 –5 –4 –3 –2 –1 1 2 3 4 5 ( 2 3, 6) ( 2 2, 1) ( 2 1, 2 2) (3, 6) (1, 2 2) (0, 2 3) Vertex y 5 x -5-4-3-2-1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 x y (-3, 9) (-2, 4) (-1, 1) (3, 9) (2, 4) (1, 1) (0, 0) Vertex 629 7.5 Graphing Quadratic Functions Here is a summary of the preceding information: EXAMPLE 1 Sketch the graph of f (x) = x 2 − 6x + 5. Solution To find the x-intercepts, we let y = 0 and solve for x. 0 = x 2 − 6x + 5 0 = (x − 5)(x − 1) x = 5 or x = 1 To find the coordinates of the vertex, we first find x = −b ___ 2a = −(−6) ______ 2(1) = 3 The x-coordinate of the vertex is 3. To find the y-coordinate, we substitute 3 for x in our original equation. f (3) = 3 2 − 6(3) + 5 = 9 − 18 + 5 = −4 The graph crosses the x-axis at 1 and 5 and has its vertex at (3, −4). Plotting these points and connecting them with a smooth curve, we have the graph shown in Figure 3. The graph is a vertical parabola that opens up, so we say the graph is concave up. The vertex is the lowest point on the graph. (Note that the graph crosses the y-axis at 5, which is the value of y we obtain when we let x = 0.) Finding the Vertex by Completing the Square Another way to locate the vertex of the parabola in Example 1 is by completing the square on the first two terms on the right side of the equation f (x) = x 2 − 6x + 5. In this case, we would do so by adding 9 to and subtracting 9 from the right side of the equation. This amounts to adding 0 to the equation, so we know we haven’t changed its solutions. PROPERTY Graphing Parabolas I The graph of the quadratic function f (x) = ax 2 + bx + c, a ≠ 0, will be a vertical parabola with 1.

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  Teaching and Learning Algebra

  • Doug French(Author)
  • 2004(Publication Date)
  • Continuum

    (Publisher)

  Graphs should also be introduced in conjunction with factorizing quadratic expressions, dis-cussed in Chapter 5, and linked to solving quadratic equations, considered in Chapter 7. The link between the factorized form of a quadratic function, the solution of the corresponding equation and the intersection of the curve and the x axis are all fairly straightforward ideas, although students do not always immediately realize why a pair of very different looking functions like y = x 2 -4x + 3 and y = (x -)(x -3) must have identical graphs. The translations which link the position of the graph of a quadratic function of the form y = x 2 + bx + c and the graph of y = x 2 require much more extensive consideration for which a graph plotter is an invaluable tool. Students have little difficulty in seeing that the graph of y = x 2 can be translated vertically up and down by adding or subtracting a constant. Moreover, a simple example like y = x 2 - 4, shown in Figure 6.7, can be linked readily to the factorized form, y = (x -2)(x + 2), and to its intersections with the ;c axis at 2 and -2 and the y axis at —4. Figure 6.7 The graph of y = x 1 -4 The problem of translating y = x 2 to left or right is much more interesting and is a good focus for class discussion with the graph displayed on a graph plotter screen for all to see. 88 Teaching and Learning Algebra T: How can we move the graph of y =* x 1 three units to the right? What form will the equation take? A: It's y -x 2 + 3. [Without comment, T displays this on the screen.] Oh no, that moves it up. B: Try y -(x + 3) 2 . [Again without comment, T deletes the previous graph and displays the new one on the screen.] Oh dear, it has moved the wrong way. T: So, what should it be? B: It must be y * (x -3) 2 . [Retaining the previous graph, T displays the new graph on the screen, which now looks like Figure 6.8.] Yes, it's right now. T: Why must it be y -(x -3) 2 ? C: Because it has to be zero when x = 3 to move it to the right.

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  Beginning Algebra

  Connecting Concepts through Applications

  • Mark Clark, Cynthia Anfinson(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ● ● A quadratic equation in two variables has the form y 5 ax 2 1 bx 1 c , where a ? 0 . ● ● A parabola is the shape of the graph of a quadratic equation in two variables. ● ● For a quadratic equation in the form y 5 ax 2 1 bx 1 c , if a . 0 (positive), the parabola opens upward, and if a , 0 (negative), the parabola opens downward. ● ● The vertex of a parabola is the lowest point on the graph for a parabola that opens upward, and it is the highest point on the graph for a parabola that opens downward. The vertex of a parabola that opens upward is called the minimum point . The vertex of a parabola that opens downward is called the maximum point . ● ● The x -coordinate of the vertex of the quadratic equation in two variables, y 5 ax 2 1 bx 1 c , is given by x 5 2 b 2 a . The y -coordinate is found by substituting the x -coordinate into the equation and simplifying. ● ● To interpret the meaning of the vertex in an application, be sure to explain what both coordinates of the vertex mean. Indicate if the vertex is a maximum or minimum point. ● ● The vertical line that passes through the vertex of a parbola is called the axis of symmetry . The axis of symmetry formula is x 5 2 b 2 a . ● ● To graph a quadratic equation in two variables , use these steps: 1. Find the equation of the axis of symmetry. 2. Find the coordinates of the vertex. 3. Graph a total of seven points. Find the values of additional symmetric points as necessary. There should be at least three points to the left and right of the vertex.

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  Precalculus

  Functions and Graphs

  • Earl Swokowski, Jeffery Cole(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  SECTION 2.6 Quadratic Functions 151 Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. EXAMPLE 1 Sketching the graph of a quadratic function Sketch the graph of f if (a) f s x d 5 2 1 2 x 2 (b) f s x d 5 2 1 2 x 2 1 4 Solution (a) Since f is even, the graph of f s that is, of y 5 2 1 2 x 2 d is symmetric with respect to the y -axis. It is similar in shape to but wider than the parabola y 5 2 x 2 , sketched in Figure 5 of Section 2.5. Several points on the graph are s 0, 0 d , s 1, 2 1 2 d , s 2, 2 2 d , and s 3, 2 9 2 d . Plotting and using symmetry, we obtain the sketch in Figure 2. (b) To find the graph of y 5 2 1 2 x 2 1 4, we shift the graph of y 5 2 1 2 x 2 upward a distance 4, obtaining the sketch in Figure 3. ■ If f s x d 5 ax 2 1 bx 1 c and b ± 0 , then, by completing the square, we can change the form to f s x d 5 a s x 2 h d 2 1 k for some real numbers h and k . This technique is illustrated in the next example. EXAMPLE 2 Expressing a quadratic function as f s x d 5 a s x 2 h d 2 1 k If f s x d 5 3 x 2 1 24 x 1 50 , express f s x d in the form a s x 2 h d 2 1 k . Solution 1 Before completing the square, it is essential that we factor out the coefficient of x 2 from the first two terms of f s x d , as follows: f s x d 5 3 x 2 1 24 x 1 50 given 5 3 s x 2 1 8 x 1 d 1 50 factor out 3 from 3 x 2 1 24 x We now complete the square for the expression x 2 1 8 x within the parentheses by adding the square of half the coefficient of x —that is, s 8 2 d 2 , or 16.

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  Mathematical Applications for the Management, Life, and Social Sciences

  • Ronald Harshbarger, James J. Reynolds(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SECTION 2.2 Quadratic Functions: Parabolas 135 The information that is useful in graphing quadratic functions is summarized as follows. Form: y 5 f (x) 5 ax 2 1 bx 1 c Graph: parabola a 7 0 parabola opens upward; vertex is a minimum point a 6 0 parabola opens downward; vertex is a maximum point Coordinates of vertex: x 5 2b 2a , y 5 f a 2b 2a b Axis of symmetry equation: x 5 2b 2a x-intercepts or zeros (if real*): x 5 2b 1 ! b 2 2 4ac 2a , x 5 2b 2 ! b 2 2 4ac 2a y-intercept: Let x 5 0; then y 5 c. -b 2a f ( ) Vertex (minimum) c Axis of symmetry -b 2a y = ax 2 + bx + c (a > 0) x y Vertex (maximum) Axis of symmetry x -b 2a f ( ) -b 2a y y = ax 2 + bx + c (a < 0) c Graphs of Quadratic Functions Graph of a Quadratic Function For the function y 5 4x 2 x 2 , determine whether its vertex is a maximum point or a min- imum point and find the coordinates of this point, find the zeros (if any exist), find the y-intercept, and sketch the graph. Solution The proper form is y 5 2x 2 1 4x 1 0, so a 5 21, b 5 4, c 5 0. Because a 6 0, the parabola opens downward and the vertex is the highest (maximum) point. The vertex occurs at x 5 2b 2a 5 24 2(21) 5 2. The y-coordinate of the vertex is f (2) 5 2(2) 2 1 4(2) 5 4. The zeros of the function are solutions to 2x 2 1 4x 5 0 x(2x 1 4) 5 0 so x 5 0 or x 5 4 The y-intercept is c 5 0 (at the origin). The graph of this function can be found by drawing a parabola with these points [see Figure 2.7(a)]. ■ *If the zeros are not real, the graph does not cross the x-axis. EXAMPLE 3 Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

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  Functions Modeling Change

  A Preparation for Calculus

  • Eric Connally, Deborah Hughes-Hallett, Andrew M. Gleason(Authors)
  • 2019(Publication Date)
  • Wiley

    (Publisher)

  CONTENTS 3.1 Introduction to the Family of Quadratic Functions . . . . . . . . . . . . . . . . . . 106 Finding the Zeros of a Quadratic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Concavity and Rates of Change for Quadratic Functions . . . . . . . . . . . . . . . . . . . 107 Finding a Formula From the Zeros and Vertical Intercept . . . . . . . . . . . . . . . . . . . . . . . . . 109 Formulas for Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Summary for Section 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 3.2 The Vertex of a Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 The Vertex Form of a Quadratic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Finding a Formula Given the Vertex and Another Point . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Modeling with Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 Summary for Section 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 STRENGTHEN YOUR UNDERSTANDING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Chapter 3 QUADRATIC FUNCTIONS 106 Chapter 3 QUADRATIC FUNCTIONS 3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS A baseball is “popped” straight up by a batter. The height of the ball above the ground is given by the function  =  () = −16 2 + 47 + 3, where  is time in seconds after the ball leaves the bat and  is in feet. See Figure 3.1. Note that the path of the ball is straight up and down, although the graph of height against time is a curve. The ball goes up fast at first and then more slowly because of gravity; thus the graph of its height as a function of time is concave down.

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  Precalculus

  Functions and Graphs, Enhanced Edition

  • Earl Swokowski, Jeffery Cole(Authors)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Typically, a given quadratic function closely resembles one of the three forms listed in the following chart. h 0 0 t 50.9 h t 0 6.3 t 43.9. t 25.1 18.8 43.9 t 25.1 25.1 6.3 h t h t 5000 Y 1 Y 1 16 x 2 803 x 600 and Y 2 5000 h t 5000 0, 60, 5 by 0, 11,000, 1000 . t 0 h 25 10,675 t b 2 a 803 2 16 25.1, FIGURE 10 [0, 60, 5] by [0, 11,000, 1000] Form Vertex ( h , k ) x -intercepts (if there are any) (1) h and k as in the form (see below) (2) (3) (see below) x b 2 a b 2 4 ac 2 a k f h h b 2 a , y f x ax 2 bx c x x 1 , x 2 k f h h x 1 x 2 2 , y f x a x x 1 x x 2 x h k a y f x a x h 2 k Relationship Between Quadratic Function Forms and Their Vertex and x -intercepts Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 160 CHAPTER 2 FUNCTIONS AND GRAPHS If the radicands in (1) or (3) are negative, then there are no x -intercepts. To find the x -intercepts with form (1), use the special quadratic equation on page 46. If you have a quadratic function in form (3) and want to find the vertex and the x -intercepts, it may be best to first find the x -intercepts by using the quadratic formula. Then you can easily obtain the x -coordinate of the vertex, h , since Of course, if the function in form (3) is easily factorable, it is not necessary to use the quadratic formula. We will discuss parabolas further in a later chapter. b 2 a b 2 4 ac 2 a h b 2 4 ac 2 a . Exer. 1–4: Find the standard equation of any parabola that has vertex V . 1 2 3 4 Exer. 5–12: Express in the form . 5 6 7 8 9 10 11 12 Exer. 13–22: (a) Use the quadratic formula to find the zeros of f .

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  College Algebra

  • R. Gustafson, Jeff Hughes(Authors)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Its lowest point, Vs0, 24d, is called the vertex. Because the y-axis divides the parabola into two congruent halves, it is called an axis of symmetry. We say that the parabola is symmetric about the y-axis. Graphing an Equation A graphing calculator can be used to draw the graph of the equation y 5 x 2 2 4. When using a graphing calculator, we must enter the equation and then set the window. ● To enter the equation, press Y= , which opens the window shown in Figure 2-47(a) and enter the right side of the equation. ● To set the window, press WINDOW as shown in Figure 2-47(b). We can experi- ment with different windows, or we can use the ZOOM menu to select some default windows. ● Press GRAPH , and the equation is graphed as shown in Figure 2-47(c). (a) (b) (c) FIGURE 2-47 Accent on Technology 2. Use Symmetry to Help Graph Equations Symmetry is a tool that we can use to help graph equations. There are many exam- ples of symmetry in the real world. There are several ways in which a graph can have symmetry. y 5 x 2 2 4 x y sx, yd 23 5 s23, 5d 22 0 s22, 0d 21 23 s21, 23d 0 24 s0, 24d 1 23 s1, 23d 2 0 s2, 0d 3 5 s3, 5d Take Note For each input value x there corre- sponds exactly one output value y. Therefore, the equation y 5 x 2 2 4 represents y as a function of x. The equation can be written using func- tional notation as f sxd = x 2 2 4. Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Chapter 2 Functions and Graphs 250 1. If the point sx, 2yd lies on the graph whenever the point sx, yd does, the graph is symmetric about the x-axis.

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  Intermediate Algebra

  Concepts and Graphs 2E

  • Charles P. McKeague(Author)
  • 2019(Publication Date)
  • XYZ Textbooks

    (Publisher)

  618 CHAPTER 7 Quadratic Functions The two points just found are (0, 1) and (2, 1). Plotting these two points along with the vertex (1, −2), we have the graph shown in Figure 4. EXAMPLE 4 Graph y = −2x 2 + 6x − 5. SOLUTION Letting y = 0, we have 0 = −2x 2 + 6x − 5 Again, the right side of this equation does not factor. The discriminant is b 2 − 4ac = 36 − 4(−2)(−5) = −4, which indicates that the solutions are complex numbers. This means that our original equation does not have x-intercepts. The graph does not cross the x-axis. Let’s find the vertex. (0, 1) -5-4-3-2-1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 x y (2, 1) (1, -2) FIGURE 4 Using our formula for the x-coordinate of the vertex, we have x = −b ___ 2a = −6 _____ 2(−2) = 6 __ 4 = 3 __ 2 To find the y-coordinate, we let x = 3 _ 2 : y = −2  3 __ 2  2 + 6  3 __ 2  − 5 = −18 ____ 4 + 18 __ 2 − 5 = −18 + 36 − 20 ____________ 4 = − 1 __ 2 Finding the vertex by completing the square is a more complicated matter. To make the coefficient of x 2 a 1, we must factor −2 from the first two terms. To complete the square inside the parentheses, we add 9 _ 4 . Since each term inside the parentheses is multiplied by −2, we add 9 _ 2 outside the parentheses so that the net result is the same as adding 0 to the right side: y = −2(x 2 − 3x + 9 __ 4 ) − 5 = −2  x 2 − 3x + 9 __ 4  − 5 + 9 __ 2 = −2  x − 3 __ 2  2 − 1 __ 2 7.5 Graphing Parabolas 619 The vertex is  3 _ 2 , − 1 _ 2  . Because this is the only point we have so far, we must find two others. Let’s let x = 3 and x = 0, because each point is the same distance from x = 3 _ 2 and on either side: When x = 3 When x = 0 y = −2(3) 2 + 6(3) − 5 y = −2(0) 2 + 6(0) − 5 = −18 + 18 − 5 = 0 + 0 − 5 = −5 = −5 The two additional points on the graph are (3, −5) and (0, −5). Figure 5 shows the graph. The graph is concave down.

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  Intermediate Algebra

  Connecting Concepts through Applications

  • Mark Clark, Cynthia Anfinson(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  The range of a quadratic function without a context starts or ends with the output value of the vertex, as this value is either the lowest or highest output value of the function. If the parabola opens upward, the range tends to infinity. If the parabola opens downward, the range tends to negative infinity. DOMAIN AND RANGE OF QUADRATIC FUNCTIONS Domain All real numbers, 1 2` , `2 Range If the vertex of the quadratic function is (h, k), then the range is 1 2`, k 4 , or y # k when the graph opens downward 3 k, `2 , or y $ k when the graph opens upward Example 5 Domain and range of quadratic functions with no context Find the domain and range for the following functions. a. f 1 x 2 5 2.51 x 2 62 2 1 2 b. f 1 x 2 5 20.251 x 1 82 2 1 10 c. g1 x 2 5 1 x 2 32 2 2 5 Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. S E C T I O N 4 . 2 G r a p h i n g Q u a d r a t i c F u n c t i o n s f r o m V e r t e x F o r m 355 SOLUTION a. We graphed this function in Example 1. Looking at the graph, we see the graph continues to go up and gets wider as x goes to both positive and negative infinity. Output values increase to infinity Lowest output value y = 2 80 40 100 60 20 0 2 y x 6 10 4 8 12 Domain: All real numbers Range: y $ 2 This function’s domain is all real numbers. The lowest point on the parabola is the vertex, so the lowest output value for the function is y 5 2, and the outputs continue up to infinity. Thus, we get the following: Domain: All real numbers or 1 2` , `2 Range: y $ 2 or 3 2, `2 b. We graphed this function in Example 2 on pages 350–351.

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