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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**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. **8.1 Asymptotes** - Vertical asymptote: $$ x=2 $$ - Horizontal asymptote: $$ y=1 $$ **8.2 Intercepts** - $$ x $$-intercept: $$ (0,0) $$ - $$ y $$-intercept: $$ (0,0) $$ **8.3 Graph Sketch** - Plot vertical asymptote at $$ x=2 $$ and horizontal asymptote at $$ y=1 $$. - The graph has two branches: - For $$ x<2 $$: $$ h(x) $$ approaches $$-\infty$$ as $$ x $$ approaches 2 from the left. - For $$ x>2 $$: $$ h(x) $$ approaches $$+\infty$$ as $$ x $$ approaches 2 from the right. - As $$ x \to \pm\infty $$, $$ h(x) $$ approaches 1. - Plot the intercept at $$ (0,0) $$ and sketch the graph accordingly. **8.4 Domain** All real numbers except $$ x=2 $$: $$ \{x \in \mathbb{R} \mid x \neq 2\} $$ **8.5 Transformations** - **8.5.1 $$ f(x)=h(x+3) $$:** Shifts the graph of $$ h $$ left by 3 units. - New vertical asymptote: $$ x=-1 $$ - Horizontal asymptote remains $$ y=1 $$ - **8.5.2 $$ f(x)=h(x)-2 $$:** Shifts the graph of $$ h $$ down by 2 units. - New horizontal asymptote: $$ y=-1 $$ - Vertical asymptote remains $$ x=2 $$

Given:
8.1 Give the equations of the asymptotes of dind
8.2 Determine the -and -intercepts of the graph of .
8.3 Sketch the graph of using the grid on the DIAGRAM SHEET
8.4 Give the domain of .
8.5 Describe the transformation of to if:
8.5.1

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Given: 8.1 Give the equations of the asymptotes of dind 8.2 Determine the -and -intercepts of the graph of . 8.3 Sketch the graph of using the grid on the DIAGRAM SHEET 8.4 Give the domain of . 8.5 Describe the transformation of to if: 8.5.1