Given: \( f(x)=\frac{3}{x-2}+1 \) 10.1 Write down the equations of the asymptotes of \( f \). 10.2 Determine coordinates of B , the \( x \)-intercept of \( f \). 10.3 Write down the domain of \( g \) if \( g(x)=f(x+1) \). 10.4 One of the axes of symmetry of \( f \) is an increasing function. Write down the equation of this axis of symmetry. \( y=x-p+q \) \( \qquad y=x-c 2)+1 \)

The function \( f(x) = \frac{3}{x-2} + 1 \) has a vertical asymptote where the denominator is zero, which occurs at \( x = 2 \). The horizontal asymptote is determined by the behavior of the function as \( x \) approaches infinity (or negative infinity). Here, as \( x \to \pm \infty \), \( f(x) \) approaches \( 1 \). Therefore, the asymptotes are \( x = 2 \) (vertical) and \( y = 1 \) (horizontal). To find the \( x \)-intercept of the function, we set \( f(x) = 0 \): \[ 0 = \frac{3}{x-2} + 1 \implies \frac{3}{x-2} = -1 \implies 3 = - (x-2) \implies x + 1 = 0 \implies x = -1. \] This means the \( x \)-intercept, point B, has coordinates \( (-1, 0) \). Now, consider the transformation function \( g(x) = f(x+1) = \frac{3}{(x+1)-2} + 1 = \frac{3}{x-1} + 1 \). The vertical asymptote moves to \( x = 1 \), so the domain of \( g \) is all real numbers except for \( x = 1 \), or in interval notation, \( (-\infty, 1) \cup (1, \infty) \). The function \( f(x) \) is not symmetrical about the y-axis but has an axis of symmetry that is an increasing function, represented by the line through the point at which \( f(x) \) reflects over itself. The configurations of the problem imply the axis of symmetry could be expressed as \( y = x - p + q \) or in a simpler linear form where \( y=x+c \). In this setting, we want to find corresponding values for constants that make the function symmetrical without a specific numerical solution readily solvable, but typically will fall back to \( y = x \) for a basic orientation around the origin in such isomorphic mappings.