Given: \( f(x)=\frac{s}{x-2}+1 \) 10.1 Write down the equations of the asymptotes of \( f \). 10.2 Determatie 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.
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The function \( f(x) = \frac{s}{x-2} + 1 \) has a vertical asymptote whenever the denominator equals zero, which occurs at \( x = 2 \). This means that the vertical asymptote is the line \( x = 2 \). Additionally, as \( x \) approaches infinity or negative infinity, \( f(x) \) approaches 1, leading to a horizontal asymptote at \( y = 1 \). To find the \( x \)-intercept \( B \), we set \( f(x) = 0 \): \[ 0 = \frac{s}{x-2} + 1 \Rightarrow \frac{s}{x-2} = -1 \Rightarrow s = -(x - 2) \Rightarrow x = 2 + s. \] Therefore, the coordinates of the \( x \)-intercept \( B \) are \( (2+s, 0) \). For the domain of \( g(x) = f(x+1) \), we substitute \( x + 1 \) into the function, which modifies the vertical asymptote to \( x + 1 = 2 \) or \( x = 1 \). Hence, the domain of \( g \) is all real numbers except \( x = 1 \), or \( (-\infty, 1) \cup (1, \infty) \). Since \( f(x) \) has the form of a hyperbola, its axes of symmetry can be found from its behavior. The axis of symmetry for hyperbolas is often related to \( x = 2 \), but an increasing function in this case occurs around the transformation of the \( y \) values. Thus, the equation of the increasing axis of symmetry can be derived as \( y = x - 1 \).
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