\( 1 \leftarrow \quad \) Find the \( x \)-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema. \[ f(x)=\frac{x^{2}-4 x+4}{x-6} \]

To find the relative extrema of the function \( f(x)=\frac{x^{2}-4 x+4}{x-6} \), we first simplify the function. The numerator can be factored as \( (x-2)^2 \), so we have: \[ f(x) = \frac{(x-2)^2}{x-6} \] Next, we need to find the derivative \( f'(x) \) to locate the critical points where the relative extrema may occur. Using the quotient rule: \[ f'(x) = \frac{(x-6)(2(x-2)) - (x-2)^2(1)}{(x-6)^2} \] This simplifies to: \[ f'(x) = \frac{2(x-2)(x-6) - (x-2)^2}{(x-6)^2} \] Setting the numerator equal to zero to find critical points: \[ 2(x-2)(x-6) - (x-2)^2 = 0 \] Factoring out \( (x-2) \): \[ (x-2)(2(x-6) - (x-2)) = 0 \] This gives us \( x-2 = 0 \) or \( 2x - 12 - x + 2 = 0 \), leading to \( x = 2 \) or \( x = 10 \). Now we evaluate \( f \) at these critical points: 1. For \( x = 2 \): \[ f(2) = \frac{(2-2)^2}{2-6} = \frac{0}{-4} = 0. \] 2. For \( x = 10 \): \[ f(10) = \frac{(10-2)^2}{10-6} = \frac{8^2}{4} = \frac{64}{4} = 16. \] Thus, the relative extrema occur at \( x = 2 \) with a value of \( 0 \) and \( x = 10 \) with a value of \( 16 \). The function has a relative minimum at \( (2, 0) \) and a relative maximum at \( (10, 16) \).