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Published byCarol Marianna Gilbert Modified over 9 years ago
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2 Any function of the form y = f (x) = ax 2 + bx + c where a 0 is called a Quadratic Function
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3 Example: f(x) = 3x 2 – 2x – 5 a = 3, b = -2, c = -5 Note that if a = 0 we simply have the linear function f(x) = bx + c f(x) = bx + c Consider the quadratic function in general form:
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4 Consider the simplest quadratic equation y = x 2 y = x 2 Here a = 1, b = 0, c = 0 Plotting some ordered pairs (x, y) we have: y = f (x ) = x 2 y = f (x ) = x 2 x f (x ) (x, y ) -3 9 (-3, 9) -2 4 (-2, 4) -1 1 (-1, 1) 0 0 (0, 0) 1 1 (1, 1) 1 1 (1, 1) 2 4 (2, 4) 2 4 (2, 4) 3 9 (3, 9) 3 9 (3, 9)
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5 (x, y) (-3, 9) (-2, 4) (-1, 1) (0, 0) (1, 1) (2, 4) (3, 9) x y -3 -2 -1 1 2 3 4321 (-2, 4) Vertex (0, 0) (2, 4) y = x 2 A parabola with the y-axis as the axis of symmetry. (1, 1) (-1, 1)
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6 Graphs of y = ax 2 will have similar form and the value of the coefficient ‘a ’ determines the graph’s shape. x y -3 -2 -1 1 2 3 4321 y = x 2 y = 2x 2 y = 1 / 2 x 2 a > 0 opening up
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7 x y y = -2x 2 a < 0 opening down In general the quadratic term ax 2in the quadratic function f (x (x ) = ax 2 +bx +bx + c determines the way the graph opens.
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8 It turns out that the details of a quadratic function can be found by considering its coefficients a, b and c as follows: (1) Opening up (a > 0), down (a 0), down (a < 0) (2) y –intercept: c (3) x -intercepts from solution of y = ax 2 + bx + c = 0 y = ax 2 + bx + c = 0 (4) vertex = You solve by factoring or the quadratic formula
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9 Example: y = f (x ) = x 2 - x - 2 here a = 1, b = -1 and c = -2 (1) opens upwards since a > 0 (2) y –intercept: -2 (3) x -intercepts from x 2 - x - 2 = 0 or (x -2)(x +1) = 0 or (x -2)(x +1) = 0 x = 2 or x = -1 x = 2 or x = -1 (4) vertex: Plug in x = to find the y-value
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10 x y -2 -1 0 1 2 -2-3 (-1, 0) (2, 0) y = x 2 - x - 2
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11 Example: y = j (x ) = x 2 - 9 here a = 1, b = 0 and c = -9 (1) opens upwards since a > 0 (2) y –intercept: -9 (3) x -intercepts from x 2 - 9 = 0 or x 2 = 9 x = 3 or x 2 = 9 x = 3 (4) vertex at (0, -9)
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12 x y -3 0 3 -9 (-3, 0) (3, 0) y = x 2 - 9 (0, -9)
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13 Example: y = g (x ) = x 2 - 6x + 9 here a = 1, b = -6 and c = 9 (1) opens upwards since a > 0 (2) y –intercept: 9 (3) x -intercepts from x 2 - 6x + 9 = 0 or (x - 3)(x - 3) = 0 x = 3 only or (x - 3)(x - 3) = 0 x = 3 only (4) vertex:
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14 x y 3 9 (3, 0) (0, 9) y = x 2 - 6x + 9
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15 Example: y = f (x ) = -3x 2 + 6x - 4 here a = -3, b = 6 and c = -4 (1) opens downwards since a < 0 (2) y –intercept: -4 (3) x -intercepts from -3x 2 + 6x - 4 = 0 (there are no x -intercepts here) (there are no x -intercepts here) (4) vertex at (1, -1) Vertex is below x-axis, and parabola opens down!
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16 x y 1 2 -4 (1, -1) (0, -4) y = -3x 2 + 6x - 4
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17 The Quadratic Formula It is not always easy to find x -intercepts by factoring ax 2 + bx + c when solving ax 2 + bx + c = 0 Quadratic equations of this form can be solved for x using the formula:
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18 Example: Solve x 2 − 6x + 9 = 0 here a = 1, b = -6 and c = 9 as found previously Note: the expression inside the radical is called the “discriminant” Note: discriminant = 0 one solution
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19 Example: Solve x 2 - x - 2 = 0 here a = 1, b = -1 and c = -2 Note: discriminant > 0 two solutions
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20 Example: Find x -intercepts of y = x 2 - 9 Solve x 2 - 9 = 0 a = 1, b = 0, c = -9 x = 3 or x = -3 Note: discriminant > 0 two solutions
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21 Example: Find the x -intercepts of y = f (x) = -3x 2 + 6x - 4 y = f (x) = -3x 2 + 6x - 4 a = -3, b = 6 and c = -4 Solve -3x 2 + 6x - 4 = 0 there are no x -intercepts as we discovered in an earlier plot of y = -3x 2 + 6x - 4 Note: discriminant < 0 no Real solutions
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22 The end.
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