f(x)=(x−6)2−9Which 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)=x2−12x+27
Q. f(x)=(x−6)2−9Which 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)=x2−12x+27
Expand and Simplify: We are given the function f(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.Let's expand (x−6)2 first:(x−6)2=x2−2×(6)×x+62=x2−12x+36
Subtract Constant: Now, we subtract 9 from the expanded form to get the complete function:f(x)=x2−12x+36−9f(x)=x2−12x+27
Factor Quadratic: We now have the function in the form f(x)=x2−12x+27. To find the zeros, we can factor this quadratic equation. The factors of 27 that add up to −12 are −3 and −9. Therefore, we can write the function as:f(x)=(x−3)(x−9)
Match Factored Form: We check the options given to see which one matches our factored form of the function:(A) f(x)=(x−9)(x−6) does not match.(B) f(x)=(x−9)(x−3) matches our factored form.(C) f(x)=(x+3)(x+9) does not match.(D) f(x)=x2−12x+27 is the expanded form but does not show the zeros as constants or coefficients.
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