Home Forums Math Which equation is y=3(x-2)^2-(x-5)^2 rewritten in vertex form? Reply To: Which equation is y=3(x-2)^2-(x-5)^2 rewritten in vertex form?

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suseelk
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    Answer : Option D

    Explanation :
    => y = 3(x – 2)^2 – (x – 5)^2
    => y = 3(x – 2)(x – 2) – (x – 5)(x – 5)
    => y = 3(x(x – 2) – 2(x – 2)) – (x(x – 5) – 5(x – 5))
    => y = 3(x(x) – x(2) – 2(x) + 2(2)) – (x(x) – x(5) – 5(x) + 5(5))

    => y = 3(x^2 – 2x – 2x + 4) – (x^2 – 5x – 5x + 25)
    => y = 3(x^2 – 4x + 4) – (x^2 – 10x + 25)
    => y = 3(x^2) – 3(4x) + 3(4) – (x^2) + (10x) – (25)
    => y = (3x^2 – 12x + 12) + (-x^2 + 10x – 25)
    => y = (3x^2 – x^2) + (-12x + 10x) + (12 – 25)
    => y = 2x^2 – 2x – 13

    add 13 on both sides
    => y + 13 = 2x^2 – 2x + 1/2y + 13 + 1/2 = 2(x^2 – x + 1/4)
    => y + 27/2 = 2(x^2 – 1/2x – 1/2x + 1/4)
    => y + 27/2 = 2(x(x) – x(0.5) – 0.5(x) + 0.5(0.5))
    => y + 27/2 = 2(x(x – 1/2) – 1/2(x – 1/2))
    => y + 27/2 = 2(x – 1/2)(x – 1/2)
    => y + 27/2 = 2(x – 1/2)^2

    subtract 13.5 on both sides

    Hence, y = 2(x – 1/2)^2 – 27/2

    • This reply was modified 2 years, 9 months ago by ProtonsTalk.
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