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?
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 3 years, 1 month ago by
ProtonsTalk.