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f(x)=x^(2)-6x-7
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-7)(x+1)
(B) 
f(x)=(x-1)(x+7)
(c) 
f(x)=x(x-6)-7
(D) 
f(x)=(x-6)^(2)-7

f(x)=x26x7 f(x)=x^{2}-6 x-7 \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)=(x7)(x+1) f(x)=(x-7)(x+1) \newline(B) f(x)=(x1)(x+7) f(x)=(x-1)(x+7) \newline(C) f(x)=x(x6)7 f(x)=x(x-6)-7 \newline(D) f(x)=(x6)27 f(x)=(x-6)^{2}-7

Full solution

Q. f(x)=x26x7 f(x)=x^{2}-6 x-7 \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)=(x7)(x+1) f(x)=(x-7)(x+1) \newline(B) f(x)=(x1)(x+7) f(x)=(x-1)(x+7) \newline(C) f(x)=x(x6)7 f(x)=x(x-6)-7 \newline(D) f(x)=(x6)27 f(x)=(x-6)^{2}-7
  1. Find Zeros: To find an equivalent form of the function that shows the zeros, we need to factor the quadratic equation f(x)=x26x7f(x) = x^2 - 6x - 7.\newlineWe look for two numbers that multiply to 7-7 and add up to 6-6.\newlineThe numbers 7-7 and +1+1 satisfy these conditions because (7)×(+1)=7(-7) \times (+1) = -7 and (7)+(+1)=6(-7) + (+1) = -6.
  2. Factor Quadratic Equation: We use these numbers to factor the quadratic equation:\newlinef(x) = x26x7x^2 - 6x - 7\newlinef(x) = (x7)(x+1)(x - 7)(x + 1)\newlineThis is the factored form of the quadratic equation, which reveals the zeros of the function.
  3. Compare with Answer Choices: We compare the factored form with the given answer choices to find the matching one.\newline(A) f(x)=(x7)(x+1)f(x) = (x - 7)(x + 1) matches the factored form we found in Step 22.
  4. Compare with Answer Choices: We compare the factored form with the given answer choices to find the matching one.\newline(A) f(x)=(x7)(x+1)f(x) = (x - 7)(x + 1) matches the factored form we found in Step 22.We check the other answer choices to confirm that they do not match the factored form we found:\newline(B) f(x)=(x1)(x+7)f(x) = (x - 1)(x + 7) does not match because it would expand to x2+6x7x^2 + 6x - 7.\newline(C) f(x)=x(x6)7f(x) = x(x - 6) - 7 does not match because it would expand to x26x7x^2 - 6x - 7, but does not show the zeros as constants or coefficients.\newline(D) f(x)=(x6)27f(x) = (x - 6)^2 - 7 does not match because it would expand to x212x+367x^2 - 12x + 36 - 7.

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