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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

$f(x)=x^{2}-6x-7$ $\newline$ 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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Quadratic equation with complex roots

Full solution

Q.

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

Factor the quadratic equation: To find the zeros, we need to factor the quadratic equation $f(x) = x^2 - 6x - 7$ .
Identify the two numbers: We look for two numbers that multiply to $-7$ and add up to $-6$ . These numbers are $-7$ and $1$ .
Write the factored form: Now we write the factored form using these numbers: $f(x) = (x - 7)(x + 1)$ .
Check answer choices: We check the answer choices to see which one matches our factored form.
Identify correct answer: The correct answer is (A) $f(x) = (x - 7)(x + 1)$ , which matches our factored form.

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