Given: \( h(x)=\frac{2}{x-2}+1 \) 8.1 Give the equations of the asymptotes of \( h \quad x=2 \) dind \( y=1 \) 8.2 Determine the \( x \)-and \( y \)-intercepts of the graph of \( h \). 8.3 Sketch the graph of \( h \) using the grid on the DIAGRAM SHEET 8.4 Give the domain of \( h \). 8.5 Describe the transformation of \( h \) to \( f \) if: 8.5.1 \( f(x)=h(x+3) \) \( 8.5 .1 \quad f(x)=h(x)-2 \)

**8.1 Asymptotes** - Vertical asymptote: The function is undefined when the denominator is zero, i.e., when \[ x-2=0 \quad \Longrightarrow \quad x=2. \] - Horizontal asymptote: When \( x \) becomes very large (or very small), \(\frac{2}{x-2}\) tends to zero so that \[ h(x) \to 0+1 = 1. \] Thus, the horizontal asymptote is \( y=1 \). **8.2 \( x \)- and \( y \)-intercepts** - \( x \)-intercept: Set \( h(x)=0 \) and solve for \( x \): \[ \frac{2}{x-2}+1=0. \] Subtract 1 from both sides: \[ \frac{2}{x-2}=-1. \] Multiply both sides by \((x-2)\) (noting \( x\neq2 \)): \[ 2=-\left(x-2\right). \] Simplify the right-hand side: \[ 2=-x+2. \] Subtract 2 from both sides: \[ 0=-x, \] so \[ x=0. \] The \( x \)-intercept is at \( (0,0) \). - \( y \)-intercept: Substitute \( x=0 \) into \( h(x) \): \[ h(0)=\frac{2}{0-2}+1=\frac{2}{-2}+1=-1+1=0. \] The \( y \)-intercept is also at \( (0,0) \). **8.3 Sketching the Graph** - Plot the asymptotes: the vertical line \( x=2 \) and the horizontal line \( y=1 \). - The intercept of the graph is at \( (0,0) \). - The graph will have two branches separated by the vertical asymptote \( x=2 \): - For \( x<2 \): as \( x \) approaches 2 from the left, \( h(x) \) goes to \(-\infty\). - For \( x>2 \): as \( x \) approaches 2 from the right, \( h(x) \) goes to \(+\infty\). - As \( x\to\pm\infty \), \( h(x) \to 1 \). Plot these features on the coordinate grid. **8.4 Domain** The function \( h(x)=\frac{2}{x-2}+1 \) is defined for all real numbers except where the denominator is zero. Thus, the domain is: \[ \{x\in\mathbb{R} \mid x\neq 2\}. \] **8.5 Transformations** - **8.5.1 For \( f(x)=h(x+3) \):** This represents a horizontal shift of the graph of \( h \) to the left by \( 3 \) units. - The vertical asymptote shifts from \( x=2 \) to \( x+3=2 \) which gives \( x=-1 \). - The horizontal asymptote remains \( y=1 \) since horizontal shifts do not affect it. - **8.5.2 For \( f(x)=h(x)-2 \):** This represents a vertical shift of the graph of \( h \) downward by \( 2 \) units. - The horizontal asymptote shifts from \( y=1 \) to \( y=1-2=-1 \). - The vertical asymptote remains \( x=2 \) because vertical shifts do not affect it.