Quadratic Graphs

11 Key excerpts on "Quadratic 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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  Elementary and Intermediate Algebra

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

    (Publisher)

  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 . 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 1 820 Chapter 11 Quadratic Equations and Functions 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 . 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 . In this case, all we are doing is taking each y -value (output) from the basic quadratic function y = x 2 and multiplying it by a factor of a . As a result, we can change the shape and direction of the basic parabola. We illustrate how this is done in the next two examples. Graph: f ( x ) = 2 x 2 . SOLUTION Because a = 2, we take each y -coordinate from the basic parabola y = x 2 and double it. Table 2 shows how this is done for several values of x .

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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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  Algebra

  A Combined Course 2E

  • Charles P. McKeague(Author)
  • 2018(Publication Date)
  • XYZ Textbooks

    (Publisher)

  C Find an equation from its graph. OBJECTIVES parabola quadratic function vertex vertical parabola concave up concave down vertex form KEY WORDS GRAPHING QUADRATIC FUNCTIONS 12.5 Graphing Quadratic Functions Participants of parkour, also called freerunning, make their own paths of travel by overcoming whatever physical obstacle stands before them. Their intense, and sometime dangerous training requires them to climb walls, flip off ramps, leap over great distances, and land with body rolls to ease the impact on their bones and joints. The art of parkour has a French origin and has been practiced for nearly a century; however, its American popularity has only begun to grow in the past decade. Imagine a freerunner as a falling object. He has just leaped over a railing at the top of a ten-foot-high brick wall. If he is falling downward at an initial velocity of 5 feet per second, the relationship between the distance s traveled and the time t is s = 5t + 16 t 2 . Problems like this one can be found by graphing a parabola. In this section, we will continue our work with parabolas and look at several ways to graph them. A Graphing Quadratic Functions As you know, the solution set to the equation y = x 2 − 3 consists of ordered pairs. As we have shown previously, one method of graphing the solution set is to find a number of ordered pairs that satisfy the equation and to graph them. We can obtain some ordered pairs that are solutions to y = x 2 − 3 by use of a table as follows: Graphing these solutions and then connecting them with a smooth curve, we have the graph of y = x 2 − 3.

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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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  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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  Mathematical Applications for the Management, Life, and Social Sciences

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

    (Publisher)

  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). 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. 136 CHAPTER 2 Quadratic and Other Special Functions -1 1 2 3 4 2 4 x y y = 4x - x 2 Vertex (2, 4) maximum (a) -2.7 6.7 -1.5 5 (b) -2.7 6.7 -1.5 5 (c) Maximum X=2 Y=4 Figure 2.7 We can graph the function y 5 4x2x 2 , discussed in Example 3, and other quadratic functions with a graphing calculator using the same methods as those for linear func- tions. Knowing the shape of a quadratic function is useful in setting a window that gives a complete graph. Figure 2.7(b) shows a graphing calculator graph of the function. In Example 3, we used the vertex and x-intercepts to sketch the graph of a quadratic function. On the other hand, we can use a graphing calculator to graph a quadratic func- tion and to find its vertex. For example, we can find the vertex of the quadratic function in Example 3 by choosing maximum under the CALC (2nd TRACE) menu and following the prompts. Figure 2.7(c) shows the maximum point (vertex) of the graph of y 5 4x2x 2 . ■ Graphing quadratic (and other polynomial) functions with Excel requires the same steps as those for graphing linear functions. See Figure 2.8 for an Excel graph of y 5 4x2x 2 and Appendix B, Sections 2.2 and 2.4, or the Online Excel Guide for procedural details. ■ y x -4 -2 0 2 4 6 –2 2 4 6 Calculator Note Spreadsheet Note Figure 2.8 CHECKPOINT 1.

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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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  Explorations in College Algebra

  • Linda Almgren Kime, Judith Clark, Beverly K. Michael(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  8.2 Visualizing Quadratics: The Vertex Form 463 Domain and range Domain: The domain for the general quadratic y = ax 2 + bx + c is all real numbers; that is, all x in the interval (−∞, +∞). Range: If the quadratic is written in vertex form, y = a(x − h) 2 + k, it’s easy to define the range, since the maximum (or minimum) of its graph is at the vertex (h, k). So if a > 0, the parabola opens up, so the range is restricted to all y in the interval [k, +∞). a < 0, the parabola opens down, so the range is restricted to all y in the interval (−∞, k]. EXAMPLE 6 Finding the Function From Its Graph Figure 8.19 shows the graph of f (x) = 2x 2 transformed into three new parabolas. Assume that each of the three new graphs retains the overall shape of f (x). a. Estimate the coordinates of the vertex for all four parabolas. b. Use your estimates from part (a) to write equations for each new parabola in Figure 8.19 in vertex form. c. Identify the domain and range of each parabola. –10 Graph A Graph B Graph C 10 f (x) = x 2 6 –6 y x –10 10 6 –6 y x –10 10 6 –6 y x –2f (x + 3) = –2(x + 3) 2 –2f (x + 3) + 5 = –2(x + 3) 2 + 5 or g(x) f (x + 3) = (x + 3) 2 FIGURE 8.18 The graph of f(x) = x 2 in Graph A is shifted horizontally to the left three units, then in Graph B stretched vertically by a factor of 2 and reflected across the x‐axis, and finally in Graph C shifted up vertically five units to generate g(x) = −2(x + 3) 2 + 5. –10 10 f (x) g(x) h(x) k(x) 5 –5 y x FIGURE 8.19 Three transformations of f (x) = 2x 2 Solution a. Vertex for: f (x) is at (0, 0); g(x) is at (3, 4); h(x) is at (3, −1); k(x) is at (−1, −4) b. g(x) = 2(x − 3) 2 + 4; h(x) = −2(x − 3) 2 − 1; k(x) = −2(x + 1) 2 − 4 c. The domain for all four functions is (−∞, +∞). The range for: f (x) is [0, +∞); g(x) is [4, +∞); h(x) is (−∞, −1]; k(x) is (−∞, −4]. 464 CHAPTER 8 Quadratics and the Mathematics of Motion Algebra Aerobics 8.2b 6. Create new functions by performing the following transformations on f (x) = x 2 .

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  Precalculus

  Functions and Graphs

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

    (Publisher)

  Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. If f s x d 5 ax 2 1 bx 1 c , then, by completing the square as in Example 2, we see that the graph of f is the same as the graph of an equation of the form y 5 a s x 2 h d 2 1 k . The graph of this equation can be obtained from the graph of y 5 ax 2 shown in Figure 4(a) by means of a horizontal and a vertical shift, as follows. First, as in Figure 4(b), we obtain the graph of y 5 a s x 2 h d 2 by shifting the graph of y 5 ax 2 either to the left or to the right, depending on the sign of h (the fig-ure illustrates the case with h . 0 ). Next, as in Figure 4(c), we shift the graph in (b) vertically a distance u k u (the figure illustrates the case with k . 0 ). It fol-lows that the graph of a quadratic function is a parabola with a vertical axis. y x y H11005 ax 2 y x y H11005 ax 2 y H11005 a ( x H11002 h ) 2 ( h , 0) y x ( h , k ) ( h , 0) y H11005 a ( x H11002 h ) 2 y H11005 a ( x H11002 h ) 2 H11001 k FIGURE 4 (a) (b) (c) The sketch in Figure 4(c) illustrates one possible graph of the equation y 5 ax 2 1 bx 1 c . If a . 0 , the point s h , k d is the lowest point on the parabola, and the function f has a minimum value f s h d 5 k . If a , 0, the parabola opens downward, and the point s h , k d is the highest point on the parabola. In this case, the function f has a maximum value f s h d 5 k . We have obtained the following result. For convenience, we often refer to the parabola y 5 ax 2 1 bx 1 c when considering the graph of this equation. EXAMPLE 3 Finding a standard equation of a parabola Express y 5 2 x 2 2 6 x 1 4 as a standard equation of a parabola with a verti-cal axis. Find the vertex and sketch the graph. Standard Equation of a Parabola with Vertical Axis The graph of the equation y 5 a s x 2 h d 2 1 k for a ± 0 is a parabola that has vertex V s h , k d and a vertical axis. The parabola opens upward if a . 0 or downward if a , 0 .

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  Intermediate Algebra

  Concepts and Graphs 2E

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

    (Publisher)

  Because the graph crosses the x-axis when y = 0, the x-intercepts are those values of x that are solutions to the quadratic equation 0 = ax 2 + bx + c. The Vertex of a Parabola The highest or lowest point on a parabola is called the vertex. The vertex for the graph of y = ax 2 + bx + c will always occur when x = −b ___ 2a To see this, we must transform the right side of y = ax 2 + bx + c into an expression that contains x in just one of its terms. This is accomplished by completing the square on the first two terms. Here is what it looks like: y = ax 2 + bx + c = a  x 2 + b __ a x  + c = a  x 2 + b __ a x +  b __ 2a  2  + c − a  b __ 2a  2 = a  x + b __ 2a  2 + 4ac − b 2 _______ 4a It may not look like it, but this last line indicates that the vertex of the graph of y = ax 2 + bx + c has an x-coordinate of −b __ 2a . Because a, b, and c are constants, the only quantity that is varying in the last expression is the x in  x + b __ 2a  2 . Because the quantity  x + b __ 2a  2 is the square of x + b __ 2a , the smallest it will ever be is 0, and that will happen when x = −b __ 2a . FIGURE 1 –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 2 2 3 -5-4-3-2-1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 x y (-3, 6) (-2, 1) (-1, -2) (3, 6) (2, 1) (1, -2) (0, -3) Vertex y = x 2 - 3 7.5 Graphing Parabolas 615 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. Here is a summary of the preceding information. EXAMPLE 1 Sketch the graph of y = 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.

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