QUESTION 5 Given: \( h(x)=\frac{2}{x-2}+1 \) \( 5.1 \quad \) Give the equations of the asymptotes of \( h \) \( 5.2 \quad \) Determine the \( x \) - and \( y \)-intercepts of the graph of \( h \). \( \begin{array}{ll}5.3 & \text { Sketch the graph of } h . \\ 5.4 & \text { Give the domain of } h . \\ 5.5 & \text { Describe the transformation of } h \text { to } f \text { if : } \\ \text { 5.5.1 } \quad f(x)=h(x+3) \\ 5.5 .1 \quad f(x)=h(x)-2\end{array} \)

### 5.1 Asymptotes of \( h(x) = \frac{2}{x-2} + 1 \) - **Vertical Asymptote:** The vertical asymptote occurs when the denominator is zero. Set \[ x - 2 = 0 \quad \Rightarrow \quad x = 2. \] - **Horizontal Asymptote:** As \( x \) approaches \(\pm \infty\), the term \(\frac{2}{x-2}\) tends to 0. Thus, \[ y = 1. \] --- ### 5.2 \( x \)- and \( y \)-Intercepts of the Graph of \( h \) - **\( y \)-Intercept:** Substitute \( x = 0 \) into \( h(x) \): \[ h(0) = \frac{2}{0-2} + 1 = \frac{2}{-2} + 1 = -1 + 1 = 0. \] So, the \( y \)-intercept is \((0, 0)\). - **\( x \)-Intercept:** Set \( h(x) = 0 \): \[ \frac{2}{x-2} + 1 = 0 \quad \Rightarrow \quad \frac{2}{x-2} = -1. \] Multiply both sides by \( x-2 \): \[ 2 = -1 \cdot (x-2) \quad \Rightarrow \quad 2 = -x + 2. \] Solving for \( x \): \[ 2 - 2 = -x \quad \Rightarrow \quad 0 = -x \quad \Rightarrow \quad x = 0. \] So, the \( x \)-intercept is also \((0, 0)\). --- ### 5.3 Sketch of the Graph of \( h \) - The vertical asymptote is the line \( x = 2 \). - The horizontal asymptote is the line \( y = 1 \). - The graph crosses the origin \((0, 0)\). - For \( x > 2 \), the function will be above or below the horizontal asymptote depending on the sign of the fraction. - For \( x < 2 \), the function will exhibit the opposite behavior. A rough sketch should include these asymptotes, the intercept at \((0,0)\), and the characteristic branches of a rational function approaching the asymptotes. --- ### 5.4 Domain of \( h \) The function \( h(x) \) is undefined where the denominator is zero: \[ x - 2 \neq 0 \quad \Rightarrow \quad x \neq 2. \] Thus, the domain is: \[ \{ x \in \mathbb{R} : x \neq 2 \}. \] --- ### 5.5 Transformations of \( h \) to \( f \) #### 5.5.1 For \( f(x) = h(x+3) \) - The transformation \( x \to x+3 \) represents a horizontal shift. - Since the input \( x \) is replaced by \( x+3 \), the graph of \( h \) shifts **3 units to the left**. #### 5.5.2 For \( f(x) = h(x) - 2 \) - The transformation \( h(x) - 2 \) represents a vertical shift. - The graph of \( h \) shifts **2 units downward**.