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f(x)=(x-6)^(2)-9
Which of the following is an equivalent form of the function 
f in which the zeros of 
f appear as constants or coefficients?
Choose 1 answer:
(A) 
f(x)=(x-9)(x-6)
(B) 
f(x)=(x-9)(x-3)
(c) 
f(x)=(x+3)(x+9)
(D) 
f(x)=x^(2)-12 x+27

f(x)=(x6)29 f(x)=(x-6)^{2}-9 \newlineWhich of the following is an equivalent form of the function f f in which the zeros of f f appear as constants or coefficients?\newlineChoose 11 answer:\newline(A) f(x)=(x9)(x6) f(x)=(x-9)(x-6) \newline(B) f(x)=(x9)(x3) f(x)=(x-9)(x-3) \newline(C) f(x)=(x+3)(x+9) f(x)=(x+3)(x+9) \newline(D) f(x)=x212x+27 f(x)=x^{2}-12 x+27

Full solution

Q. f(x)=(x6)29 f(x)=(x-6)^{2}-9 \newlineWhich of the following is an equivalent form of the function f f in which the zeros of f f appear as constants or coefficients?\newlineChoose 11 answer:\newline(A) f(x)=(x9)(x6) f(x)=(x-9)(x-6) \newline(B) f(x)=(x9)(x3) f(x)=(x-9)(x-3) \newline(C) f(x)=(x+3)(x+9) f(x)=(x+3)(x+9) \newline(D) f(x)=x212x+27 f(x)=x^{2}-12 x+27
  1. Expand and Simplify: We are given the function f(x)=(x6)29f(x)=(x-6)^{2}-9. To find an equivalent form where the zeros are visible, we need to expand the squared term and then simplify the expression.\newlineLet's expand (x6)2(x-6)^{2} first:\newline(x6)2=x22×(6)×x+62=x212x+36(x-6)^{2} = x^{2} - 2\times(6)\times x + 6^{2} = x^{2} - 12x + 36
  2. Subtract Constant: Now, we subtract 99 from the expanded form to get the complete function:\newlinef(x)=x212x+369f(x) = x^{2} - 12x + 36 - 9\newlinef(x)=x212x+27f(x) = x^{2} - 12x + 27
  3. Factor Quadratic: We now have the function in the form f(x)=x212x+27f(x) = x^{2} - 12x + 27. To find the zeros, we can factor this quadratic equation. The factors of 2727 that add up to 12-12 are 3-3 and 9-9. Therefore, we can write the function as:\newlinef(x)=(x3)(x9)f(x) = (x - 3)(x - 9)
  4. Match Factored Form: We check the options given to see which one matches our factored form of the function:\newline(A) f(x)=(x9)(x6)f(x)=(x-9)(x-6) does not match.\newline(B) f(x)=(x9)(x3)f(x)=(x-9)(x-3) matches our factored form.\newline(C) f(x)=(x+3)(x+9)f(x)=(x+3)(x+9) does not match.\newline(D) f(x)=x212x+27f(x)=x^{2}-12x+27 is the expanded form but does not show the zeros as constants or coefficients.

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