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f(x)=-(x-4)(x-2) 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? Choose 1 answer: (A) f(x)=-x^(2)+6x-8 (B) f(x)=x^(2)-6x+8 (c) f(x)=-(x-3)^(2)+1 (D) f(x)=(x+3)^(2)+1

f(x)=−(x−4)(x−2) f(x)=-(x-4)(x-2) \newlineWhich of the following is an equivalent form of the function f f in which the maximum value of f f appears as a constant or coefficient?\newlineChoose 11 answer:\newline(A) f(x)=−x2+6x−8 f(x)=-x^{2}+6 x-8 \newline(B) f(x)=x2−6x+8 f(x)=x^{2}-6 x+8 \newline(C) f(x)=−(x−3)2+1 f(x)=-(x-3)^{2}+1 \newline(D) f(x)=(x+3)2+1 f(x)=(x+3)^{2}+1

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Q. f(x)=−(x−4)(x−2) f(x)=-(x-4)(x-2) \newlineWhich of the following is an equivalent form of the function f f in which the maximum value of f f appears as a constant or coefficient?\newlineChoose 11 answer:\newline(A) f(x)=−x2+6x−8 f(x)=-x^{2}+6 x-8 \newline(B) f(x)=x2−6x+8 f(x)=x^{2}-6 x+8 \newline(C) f(x)=−(x−3)2+1 f(x)=-(x-3)^{2}+1 \newline(D) f(x)=(x+3)2+1 f(x)=(x+3)^{2}+1
  1. Expand the given function: Expand the given function f(x)=−(x−4)(x−2)f(x) = -(x-4)(x-2) to find an equivalent form.\newlinef(x)=−(x2−2x−4x+8)f(x) = -(x^2 - 2x - 4x + 8)\newlinef(x)=−(x2−6x+8)f(x) = -(x^2 - 6x + 8)\newlinef(x)=−x2+6x−8f(x) = -x^2 + 6x - 8\newlineThis is the expanded form of the quadratic function.
  2. Check the options: Check the given options to see which one matches the expanded form. Option (A) f(x)=−x2+6x−8f(x) = -x^2 + 6x - 8 matches the expanded form we found in Step 11.
  3. Determine the maximum value: Determine if the expanded form shows the maximum value as a constant or coefficient.\newlineThe function f(x)=−x2+6x−8f(x) = -x^2 + 6x - 8 is a downward-opening parabola, so the maximum value occurs at the vertex of the parabola.\newlineThe vertex form of a parabola is f(x)=a(x−h)2+kf(x) = a(x-h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.\newlineTo find the vertex, we can use the formula h=−b2ah = -\frac{b}{2a} for a quadratic function ax2+bx+cax^2 + bx + c.
  4. Calculate the vertex: Calculate the vertex using the formula h=−b2ah = -\frac{b}{2a}. For the function f(x)=−x2+6x−8f(x) = -x^2 + 6x - 8, a=−1a = -1 and b=6b = 6. h=−b2a=−62∗(−1)=−6−2=3h = -\frac{b}{2a} = -\frac{6}{2*(-1)} = -\frac{6}{-2} = 3 Now we have the x-coordinate of the vertex, which is x=3x = 3.
  5. Substitute x=3x = 3: Substitute x=3x = 3 into the function to find the yy-coordinate of the vertex, which is the maximum value of ff.f(3)=−(32)+6⋅3−8f(3) = -(3^2) + 6\cdot3 - 8f(3)=−(9)+18−8f(3) = -(9) + 18 - 8f(3)=−9+18−8f(3) = -9 + 18 - 8f(3)=9−8f(3) = 9 - 8f(3)=1f(3) = 1The vertex is at (3,1)(3, 1), so the maximum value of ff is 11.
  6. Write in vertex form: Write the function in vertex form using the vertex (3,1)(3, 1). \newlinef(x)=a(x−h)2+kf(x) = a(x-h)^2 + k\newlinef(x)=−(x−3)2+1f(x) = -(x-3)^2 + 1\newlineThis form shows the maximum value as a constant (1)(1).
  7. Match with options: Match the vertex form with the given options.\newlineOption (C) f(x)=−(x−3)2+1f(x) = -(x-3)^2 + 1 matches the vertex form we found in Step 66.

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