D) If \( f(x)=(x-2)^{2} \); find the following. \( \begin{array}{ll}\text { 1) }: 4 f(1) \times 6 f(0)= \\ \begin{array}{ll}\text { 3) }-7 f(7)+f(-8)= & \text { 2) }-8 f(2)+3 f(4)= \\ \text { 4) } \frac{-3 f(-1)}{f(3)} & \end{array}\end{array} \)

To find the values for the given expressions with \( f(x)=(x-2)^{2} \), let’s calculate \( f(x) \) at the specified points first: 1. \( f(1) = (1 - 2)^{2} = 1^{2} = 1 \) \( f(0) = (0 - 2)^{2} = (-2)^{2} = 4 \) So, \( 4 f(1) \times 6 f(0) = 4 \times 1 \times 6 \times 4 = 96 \). 2. \( f(2) = (2 - 2)^{2} = 0 \) \( f(4) = (4 - 2)^{2} = 2^{2} = 4 \) Thus, \( -8 f(2) + 3 f(4) = -8 \times 0 + 3 \times 4 = 12 \). 3. \( f(7) = (7 - 2)^{2} = 5^{2} = 25 \) \( f(-8) = (-8 - 2)^{2} = (-10)^{2} = 100 \) Therefore, \( -7 f(7) + f(-8) = -7 \times 25 + 100 = -175 + 100 = -75 \). 4. \( f(-1) = (-1 - 2)^{2} = (-3)^{2} = 9 \) \( f(3) = (3 - 2)^{2} = 1^{2} = 1 \) This gives us \( \frac{-3 f(-1)}{f(3)} = \frac{-3 \times 9}{1} = -27 \). Now let's summarize these findings: 1) \( 4 f(1) \times 6 f(0) = 96 \) 2) \( -8 f(2) + 3 f(4) = 12 \) 3) \( -7 f(7) + f(-8) = -75 \) 4) \( \frac{-3 f(-1)}{f(3)} = -27 \)