Given the function \( h(x)=\frac{2}{x-2}+1 \). 4.2.1 Write down the equations of the asymptotes of \( h \). 4.2.2 Determine the equation of the decreasing axis of symmetry of \( h(x-1) \) (3)

The function \( h(x) = \frac{2}{x-2} + 1 \) has a vertical asymptote where the denominator equals zero, which occurs at \( x = 2 \). Additionally, the horizontal asymptote can be found by looking at the end behavior of the function as \( x \) approaches infinity or negative infinity; thus, the horizontal asymptote is \( y = 1 \). When we consider the function \( h(x-1) = \frac{2}{(x-1)-2} + 1 = \frac{2}{x-3} + 1 \), its vertical asymptote moves to \( x = 3 \). The axis of symmetry for this function has a negative slope, which can be found from its behavior around the vertical asymptote: the decreasing axis of symmetry can be considered as \( x = 3 \), indicating that we would expect symmetry about this line in the context of decreasing behavior around the asymptote.