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f(x)=x^(2)-4x-1
Which of the following is an equivalent form of the function
f in which the minimum value of
f appears as a constant or coefficient?
Choose 1 answer:
(A)
f(x)=-(x-2)^(2)-5
(B)
f(x)=(x-4)(x-1)
(C)
f(x)=(x-2)^(2)-5
(D)
f(x)=x(x-4)-1

$f(x)=x^{2}-4 x-1$ $\newline$ Which of the following is an equivalent form of the function $f$ in which the minimum value of $f$ appears as a constant or coefficient? $\newline$ Choose $1$ answer: $\newline$ (A) $f(x)=-(x-2)^{2}-5$ $\newline$ (B) $f(x)=(x-4)(x-1)$ $\newline$ (C) $f(x)=(x-2)^{2}-5$ $\newline$ (D) $f(x)=x(x-4)-1$

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Compare linear and exponential growth

Full solution

Q.

f(x)=x^{2}-4 x-1

\newline

Which of the following is an equivalent form of the function

f

in which the minimum value of

f

appears as a constant or coefficient?

\newline

Choose

1

answer:

\newline

(A)

f(x)=-(x-2)^{2}-5

\newline

(B)

f(x)=(x-4)(x-1)

\newline

(C)

f(x)=(x-2)^{2}-5

\newline

(D)

f(x)=x(x-4)-1

Completing the Square: We need to complete the square to rewrite the function in vertex form, which will reveal the minimum value of the function. $\newline$ $f(x) = x^2 - 4x - 1$ $\newline$ To complete the square, we take the coefficient of $x$ , which is $-4$ , divide it by $2$ , and square it. This gives us $(-4/2)^2 = 4$ . $\newline$ We add and subtract this value inside the function to complete the square. $\newline$ $f(x) = (x^2 - 4x + 4) - 4 - 1$
Rewriting the Function: Now we can rewrite the function as a perfect square trinomial minus the constants we've added and subtracted. $\newline$ $f(x) = (x - 2)^2 - 5$ $\newline$ This is the vertex form of the quadratic function, where the vertex $(h, k)$ represents the minimum point if the parabola opens upwards (which it does since the coefficient of the $x^2$ term is positive).
Comparing with Answer Choices: We compare the vertex form we found with the given answer choices to find the matching form. $\newline$ (A) $f(x) = -(x - 2)^2 - 5$ (This is incorrect because it has a negative sign before the squared term, which would indicate a maximum value, not a minimum.) $\newline$ (B) $f(x) = (x - 4)(x - 1)$ (This is factored form, not vertex form.) $\newline$ (C) $f(x) = (x - 2)^2 - 5$ (This matches the form we found and shows the minimum value as a constant.) $\newline$ (D) $f(x) = x(x - 4) - 1$ (This is expanded form, not vertex form.)

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