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Tagged: quadratic equation, vertex form

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- July 21, 2021 at 9:31 pm #10834
a. y = 3(x – 7/2)^2 – 27/4

b. y = 2(x – 1/2)^2 – 11

c. y = 2(x – 1/2)^2 – 53/4

d. y = 2(x – 1/2)^2 – 27/2July 21, 2021 at 9:41 pm #10838::**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 – 13add 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)^2subtract 13.5 on both sides

Hence, y = 2(x – 1/2)^2 – 27/2

- This reply was modified 2 years, 10 months ago by ProtonsTalk.

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