Chapter 3

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Quadratic Functions

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Chapter 3

  1. 1. CHAPTER 3QUADRATICFUNCTIONS
  2. 2. Let a, b, and c be real numbers a ≠ 0. The function f (x) = ax2 + bx + cis called a quadratic function.The graph of a quadratic function is aparabola.Every parabola is symmetrical about a line called the axis(of symmetry). yThe intersection point of theparabola and the axis iscalled the vertex of the f (x) = ax2 + bx + cparabola. vertex x axis 2
  3. 3. The leading coefficient of ax2 + bx + c is a. y a>0When the leading coefficient f(x) = ax2 + bx + c opensis positive, the parabola upwardopens upward and the xvertex is a minimum. vertex minimum y xWhen the leading vertex maximum f(x) = ax2 + bx + ccoefficient is negative,the parabola opens downward a<0 opensand the vertex is a maximum. downward 3
  4. 4. The simplest quadratic functions are of the form f (x) = ax2 (a ≠ 0)These are most easily graphed by comparing them with thegraph of y = x2.Example: Compare the graphs of 1 y = x 2, f ( x) = x 2 and g ( x) = 2 x 2 2 y = x2 y 1 2 5 g ( x) = 2 x 2 f ( x) = x 2 x -5 5 4
  5. 5. Example: Graph f (x) = (x – 3)2 + 2 and find the vertex and axis. f (x) = (x – 3)2 + 2 is the same shape as the graph of g (x) = (x – 3)2 shifted upwards two units. g (x) = (x – 3)2 is the same shape as y = x2 shifted to the right three units. y f (x) = (x – 3)2 + 2 y = x2 g (x) = (x – 3)2 4 (3, 2) vertex x -4 4 5
  6. 6. The standard form for the equation of a quadratic function is: f (x) = a(x – h)2 + k (a ≠ 0)The graph is a parabola opening upward if a > 0 and openingdownward if a < 0. The axis is x = h, and the vertex is (h, k).Example: Graph the parabola f (x) = 2x2 + 4x – 1 and find the axis and vertex. y 2 f (x) = 2x + 4x – 1f (x) = 2x2 + 4x – 1 original equationf (x) = 2( x2 + 2x) – 1 factor out 2f (x) = 2( x2 + 2x + 1) – 1 – 2 complete the square xf (x) = 2( x + 1)2 – 3 standard forma > 0 → parabola opens upward like y = 2x2.h = –1, k = –3 → axis x = –1, vertex (–1, –3). x = –1 (–1, –3) 6
  7. 7. Example: Graph and find the vertex and x-intercepts of f (x) = –x2 + 6x + 7.f (x) = – x2 + 6x + 7 original equation y (3, 16)f (x) = – ( x2 – 6x) + 7 factor out –1f (x) = – ( x2 – 6x + 9) + 7 + 9 complete the squaref (x) = – ( x – 3)2 + 16 standard forma < 0 → parabola opens downward.h = 3, k = 16 → axis x = 3, vertex (3, 16). 4Find the x-intercepts by solving (–1, 0) (7, 0) x–x2 + 6x + 7 = 0.(–x + 7 )( x + 1) = 0 factor 4x = 7, x = –1 x=3 f(x) = –x2 + 6x + 7x-intercepts (7, 0), (–1, 0) 7
  8. 8. Example: Find an equation for the parabola with vertex (2, –1) passing through the point (0, 1). y y = f(x) (0, 1) x (2, –1)f (x) = a(x – h)2 + k standard formf (x) = a(x – 2)2 + (–1) vertex (2, –1) = (h, k)Since (0, 1) is a point on the parabola: f (0) = a(0 – 2)2 – 1 1 1 = 4a –1 and a = 1 1 2 2f ( x ) = ( x − 2) − 1 → f ( x ) = x − 2 x + 1 2 2 2 8
  9. 9. Vertex of a Parabola The vertex of the graph of f (x) = ax2 + bx + c (a ≠ 0)  b  b  is  − , f −   2a  2a  Example: Find the vertex of the graph of f (x) = x2 – 10x + 22. f (x) = x2 – 10x + 22 original equation a = 1, b = –10, c = 22 − b − 10 At the vertex, x = = =5 2a 2(1)  −b  = f (5) = 5 − 10(5) + 22 = −3 2 f  2a  So, the vertex is (5, -3). 9
  10. 10. Example: A basketball is thrown from the free throw line froma height of six feet. What is the maximum height of the ball ifthe path of the ball is: 1 2 y = − x + 2 x + 6. 9 The path is a parabola opening downward. The maximum height occurs at the vertex. −1 2 −1 y= x + 2x + 6 → a = , b=2 9 9 −b At the vertex, x = = 9. 2a  −b f  = f ( 9 ) = 15  2a  So, the vertex is (9, 15). The maximum height of the ball is 15 feet. 10
  11. 11. Example: A fence is to be built to form arectangular corral along the side of a barn barn65 feet long. If 120 feet of fencing areavailable, what are the dimensions of the x corral xcorral of maximum area? 120 – 2xLet x represent the width of the corral and 120 – 2x the length. Area = A(x) = (120 – 2x) x = –2x2 + 120 xThe graph is a parabola and opens downward. −bThe maximum occurs at the vertex where x = , − b − 120 2a a = –2 and b = 120 → x = = = 30. 2a −4 120 – 2x = 120 – 2(30) = 60The maximum area occurs when the width is 30 feet and thelength is 60 feet. 11

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