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Determining the equation of a parabola

Determining the equation of a parabola. If you are given certain points that define a parabola, you can determine its equation. If one of those points happens to be the vertex, you will only need one other point.

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Determining the equation of a parabola

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  1. Determining the equation of a parabola

  2. If you are given certain points that define a parabola, you can determine its equation. If one of those points happens to be the vertex, you will only need one other point. Determine the equation of the quadratic function, f, that passes through point (4,7) and whose vertex is (3,2). Use standard form to find the rule of correspondence. f(x) = a(x – h)2 + k First substitute vertex for h & k. To find ‘a’, substitute the given ordered pair for x & y. y = a(x – 3)2 + 2 7 = a(4 – 3)2 + 2 7 – 2 = a(1) a = 5 y = 5(x – 3)2 + 2 … Standard form y = 5(x2 – 6x + 9) + 2 y = 5x2 – 30x + 45 + 2 f(x) = 5x2 – 30x + 47 … General form

  3. #2 page 10.44 Determine the equation of the quadratic function, f, that passes through point (1,2) and whose vertex is (3,3). Use standard form to find the rule of correspondence. f(x) = a(x – h)2 + k First substitute vertex for h & k. To find ‘a’, substitute the given ordered pair for x & y. y = a(x – 3)2 + 3 2 = a(1 – 3)2 + 3 2 – 3 = a(-2)2 -1 = 4a a = -0.25 y = -0.25 (x – 3)2 + 3 … Stan. frm y = -0.25(x2 – 6x + 9) + 3 y = -0.25 x2 + 1.5x – 2.25 + 3 f(x) = -0.25x2 + 1.5x + 0.75 … Gen. frm Complete Exercise 10.10 pages 10.44, 10.45

  4. The other way that you could have to determine the rule of correspondence is if you are given the 2 zeros and any third point on the parabola. Determine the rule of correspondence of the quadratic function, f, that passes through point (0,9) and whose zeros are -1 and 3. Use this template to determine the equation: y = a(x – x1)(x – x2) where x1 & x2 are the zeros. First substitute the zeros y = a(x – (-1))(x – 3) y = a(x + 1)(x – 3) y = a(x2 – 3x + x – 3) y = a(x2 – 2x – 3) Substitute the other point 9 = a(02 – 2(0) – 3) 9 = a(-3) 3a = -9 a = -3

  5. Write the rule of correspondence with the newly found value of ‘a’ f(x) = -3(x2 – 2x – 3) f(x) = -3x2 + 6x + 9 … General Form f(x) = -3(x2 – 2x + 1) + 9 + 3 f(x) = -3(x – 1)2 + 12 … Standard Form Complete Exercise 10.11 pages 10.46, 10.47 AND 10.2 Practice Exercises 10.49 – 10.61

  6. Determine the general form of the quadratic function, with zeros (-2,0), (5,0) and point (1,-36). y = a(x – x1)(x – x2) where x1 & x2 are the zeros. First substitute the zeros y = a(x – (-2))(x – 5) y = a(x + 2)(x – 5) y = a(x2 – 5x + 2x – 10) y = a(x2 – 3x – 10) Substitute the other point -36 = a(12 – 3(1) – 10) -36 = a(-12) 12a = 36 a = 3

  7. Write the rule of correspondence with the newly found value of ‘a’ f(x) = 3(x2 – 3x – 10) f(x) = 3x2 – 9x – 30 … General Form f(x) = 3(x2 – 3x + 2.25) – 30 – 6.75 f(x) = -3(x – 1.5)2 – 36.75 … Standard Form Complete Exercise 10.11 pages 10.46, 10.47 AND 10.2 Practice Exercises 10.49 – 10.61

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