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Writing equations for parabolas?

I have to write the equation for a parabola with a y intercept of 10, x intercept of 2, and equation of axis of symmetry of x-3=0. Any help?

3 Answers

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  • 9 years ago
    Favorite Answer

    Assume that the parabola is in standard form: y = ax^2 + bx + c.

    (0,10) is on the parabola implies: 10 = a(0^2) + b(0) + c, or c = 10

    (2,0) is on the parabola implies: 0= a(2^2) + 2b + c, or 0 = 4a + 2b + 10, or 2a + b = -5

    We also know the axis of symmetry runs through an x-coordinate of -b/2a = 3, or b = -6a

    2a + b = -5

    2a - 6a = -5

    -4a = -5

    a = 5/4

    2(5/4) + b = -5

    (5/2) + b = -5

    b = -5 - (5/2) = (-10/2) - (5/2) = -15/2

    y = (5/4)x^2 - (15/2)x + 10 <== answer

  • Josh K
    Lv 6
    9 years ago

    Line of symmetry is x = 3, and an x intercept is 2, then another x intercept must be x = 4

    3-1 and 3+1

    Then y = c(x-2)(x-4)

    Now sub in the y intercept of (0,10) to find c

    10 = c*-2*-4 = 8c

    c = 5/4

    y = 5/4(x-2)(x-4)

    I hope this helps

  • 9 years ago

    x1 = 2

    h = 3

    y int. = 10, so

    c = 10

    x2 = h + (h - x1)

    x2 = 3 + (3 - 2)

    x2 = 3 + 1

    x2 = 4

    Roots: (2, 0) & (4, 0)

    y = a (x - 2)(x - 4)

    y = a (x² - 6x + 8)

    y = ax² - 6ax + 8a

    Since c = 10 and

    8a = c, then

    8a = 10, so

    a = 10 / 8

    a = 5/4

    Equation:

    y = 5/4 x² - 6(5/4)x + 8(5/4)

    y = 5/4 x² - 30/4 x + 40/4

    y = 5/4 x² - 15/2 x + 10

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    Source(s): 4/14/12
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