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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

$f(x)=x^{2}-6 x-7$ $\newline$ Which of the following is an equivalent form of the function $f$ in which the zeros of $f$ appear as constants or coefficients? $\newline$ Choose $1$ answer: $\newline$ (A) $f(x)=(x-7)(x+1)$ $\newline$ (B) $f(x)=(x-1)(x+7)$ $\newline$ (C) $f(x)=x(x-6)-7$ $\newline$ (D) $f(x)=(x-6)^{2}-7$

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Compare linear and exponential growth

Full solution

Q.

f(x)=x^{2}-6 x-7

\newline

Which of the following is an equivalent form of the function

f

in which the zeros of

f

appear as constants or coefficients?

\newline

Choose

1

answer:

\newline

(A)

f(x)=(x-7)(x+1)

\newline

(B)

f(x)=(x-1)(x+7)

\newline

(C)

f(x)=x(x-6)-7

\newline

(D)

f(x)=(x-6)^{2}-7

Find Zeros: To find an equivalent form of the function that shows the zeros, we need to factor the quadratic equation $f(x) = x^2 - 6x - 7$ . $\newline$ We look for two numbers that multiply to $-7$ and add up to $-6$ . $\newline$ The numbers $-7$ and $+1$ satisfy these conditions because $(-7) \times (+1) = -7$ and $(-7) + (+1) = -6$ .
Factor Quadratic Equation: We use these numbers to factor the quadratic equation: $\newline$ f(x) = $x^2 - 6x - 7$ $\newline$ f(x) = $(x - 7)(x + 1)$ $\newline$ This is the factored form of the quadratic equation, which reveals the zeros of the function.
Compare with Answer Choices: We compare the factored form with the given answer choices to find the matching one. $\newline$ (A) $f(x) = (x - 7)(x + 1)$ matches the factored form we found in Step $2$ .
Compare with Answer Choices: We compare the factored form with the given answer choices to find the matching one. $\newline$ (A) $f(x) = (x - 7)(x + 1)$ matches the factored form we found in Step $2$ .We check the other answer choices to confirm that they do not match the factored form we found: $\newline$ (B) $f(x) = (x - 1)(x + 7)$ does not match because it would expand to $x^2 + 6x - 7$ . $\newline$ (C) $f(x) = x(x - 6) - 7$ does not match because it would expand to $x^2 - 6x - 7$ , but does not show the zeros as constants or coefficients. $\newline$ (D) $f(x) = (x - 6)^2 - 7$ does not match because it would expand to $x^2 - 12x + 36 - 7$ .

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