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 $$ f(x)=h(x+3) $$ $$ 8.5 .1 \quad f(x)=h(x)-2 $$

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$$ \textbf{8.1\ Asymptotes} $$ The function is given by $$ h(x)=\frac{2}{x-2}+1. $$ - The vertical asymptote occurs when the denominator is zero, so set $$ x-2=0\quad \Rightarrow\quad x=2. $$ - The horizontal asymptote is found by looking at the behavior as $$ x\to\pm\infty $$. Since the fraction tends to 0, the function approaches $$ y=1. $$ $$ \textbf{8.2\ Intercepts} $$ $$\underline{\textbf{x-intercept:}}$$ Set $$ h(x)=0 $$: $$ \frac{2}{x-2}+1=0. $$ Subtract 1 from both sides: $$ \frac{2}{x-2}=-1. $$ Multiply both sides by $$ x-2 $$: $$ 2=-1\cdot(x-2) \quad\Rightarrow\quad 2=-(x-2). $$ Solve for $$ x $$: $$ 2=-x+2\quad\Rightarrow\quad -x=2-2=0\quad\Rightarrow\quad x=0. $$ $$\underline{\textbf{y-intercept:}}$$ Set $$ x=0 $$ in $$ h(x) $$: $$ h(0)=\frac{2}{0-2}+1=\frac{2}{-2}+1=-1+1=0. $$ Thus, the y-intercept is also at $$ (0,0) $$. $$ \textbf{8.3\ Sketching the Graph} $$ To sketch the graph of $$ h(x)=\frac{2}{x-2}+1 $$: - Draw the vertical asymptote at $$ x=2 $$ (a dashed line). - Draw the horizontal asymptote at $$ y=1 $$ (a dashed line). - Plot the intercept at $$ (0,0) $$. - For $$ x<2 $$ (to the left of the vertical asymptote), the graph will approach the horizontal line $$ y=1 $$ as $$ x\to -\infty $$ and tend to $$ \pm\infty $$ as $$ x\to2^- $$ (in this case, since $$ \frac{2}{x-2} $$ is negative when $$ x<2 $$, the function will tend to $$-\infty$$). - For $$ x>2 $$ (to the right of the vertical asymptote), the graph similarly approaches $$ y=1 $$ as $$ x\to\infty $$ and tends to $$ \pm\infty $$ as $$ x\to2^+ $$ (here, $$ \frac{2}{x-2} $$ is positive when $$ x>2 $$, so the function will tend to $$ +\infty$$). $$ \textbf{8.4\ Domain} $$ The function $$ h(x)=\frac{2}{x-2}+1 $$ is undefined when $$ x-2=0 $$. Thus, the domain is: $$ \{x\in\Bbb R: x\neq2\}. $$ $$ \textbf{8.5\ Transformations} $$ $$\underline{\textbf{8.5.1}}$$ For $$ f(x)=h(x+3), $$ substitute $$ (x+3) $$ into $$ h $$ in place of $$ x $$. This represents a horizontal translation $$\textbf{left}$$ by 3 units. $$\underline{\textbf{8.5.2}}$$ For $$ f(x)=h(x)-2, $$ this represents a vertical translation $$\textbf{down}$$ by 2 units. **8.1 Asymptotes** - Vertical asymptote: $$ x = 2 $$ - Horizontal asymptote: $$ y = 1 $$ **8.2 Intercepts** - x-intercept: $$ (0, 0) $$ - y-intercept: $$ (0, 0) $$ **8.3 Sketching the Graph** - Draw vertical asymptote at $$ x = 2 $$ and horizontal asymptote at $$ y = 1 $$. - Plot the intercept at $$ (0, 0) $$. - For $$ x < 2 $$, the graph approaches $$ y = 1 $$ from below and tends to $$ -\infty $$ as $$ x $$ approaches 2 from the left. - For $$ x > 2 $$, the graph approaches $$ y = 1 $$ from above and tends to $$ +\infty $$ as $$ x $$ approaches 2 from the right. **8.4 Domain** All real numbers except $$ x = 2 $$. So, $$ x \neq 2 $$. **8.5 Transformations** - $$ f(x) = h(x + 3) $$: Shift the graph of $$ h(x) $$ **left** by 3 units. - $$ f(x) = h(x) - 2 $$: Shift the graph of $$ h(x) $$ **down** by 2 units.

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