$f(x)=-(x-4)(x-2)$ $\newline$ Which of the following is an equivalent form of the function $f$ in which the maximum value of $f$ appears as a constant or coefficient? $\newline$ Choose $1$ answer: $\newline$ (A) $f(x)=-x^{2}+6 x-8$ $\newline$ (B) $f(x)=x^{2}-6 x+8$ $\newline$ (C) $f(x)=-(x-3)^{2}+1$ $\newline$ (D) $f(x)=(x+3)^{2}+1$

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Algebra 1

Compare linear and exponential growth

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f(x)=-(x-4)(x-2)

\newline

Which of the following is an equivalent form of the function

f

in which the maximum value of

f

appears as a constant or coefficient?

\newline

Choose

1

answer:

\newline

(A)

f(x)=-x^{2}+6 x-8

\newline

(B)

f(x)=x^{2}-6 x+8

\newline

(C)

f(x)=-(x-3)^{2}+1

\newline

(D)

f(x)=(x+3)^{2}+1

Expand the given function: Expand the given function $f(x) = -(x-4)(x-2)$ to find an equivalent form. $\newline$ $f(x) = -(x^2 - 2x - 4x + 8)$ $\newline$ $f(x) = -(x^2 - 6x + 8)$ $\newline$ $f(x) = -x^2 + 6x - 8$ $\newline$ This is the expanded form of the quadratic function.
Check the options: Check the given options to see which one matches the expanded form. Option (A) $f(x) = -x^2 + 6x - 8$ matches the expanded form we found in Step $1$ .
Determine the maximum value: Determine if the expanded form shows the maximum value as a constant or coefficient. $\newline$ The function $f(x) = -x^2 + 6x - 8$ is a downward-opening parabola, so the maximum value occurs at the vertex of the parabola. $\newline$ The vertex form of a parabola is $f(x) = a(x-h)^2 + k$ , where $(h, k)$ is the vertex of the parabola. $\newline$ To find the vertex, we can use the formula $h = -\frac{b}{2a}$ for a quadratic function $ax^2 + bx + c$ .
Calculate the vertex: Calculate the vertex using the formula $h = -\frac{b}{2a}$ . For the function $f(x) = -x^2 + 6x - 8$ , $a = -1$ and $b = 6$ . $h = -\frac{b}{2a} = -\frac{6}{2*(-1)} = -\frac{6}{-2} = 3$ Now we have the x-coordinate of the vertex, which is $x = 3$ .
Substitute $x = 3$ : Substitute $x = 3$ into the function to find the $y$ -coordinate of the vertex, which is the maximum value of $f$ . $f(3) = -(3^2) + 6\cdot3 - 8$ $f(3) = -(9) + 18 - 8$ $f(3) = -9 + 18 - 8$ $f(3) = 9 - 8$ $f(3) = 1$ The vertex is at $(3, 1)$ , so the maximum value of $f$ is $1$ .
Write in vertex form: Write the function in vertex form using the vertex $(3, 1)$ . $\newline$ $f(x) = a(x-h)^2 + k$ $\newline$ $f(x) = -(x-3)^2 + 1$ $\newline$ This form shows the maximum value as a constant $(1)$ .
Match with options: Match the vertex form with the given options. $\newline$ Option (C) $f(x) = -(x-3)^2 + 1$ matches the vertex form we found in Step $6$ .