QUESTION 5 Given the hyperbola: $$ f(x)=\frac{3}{x-2}+1 $$ 5.1 Write down the equations of the asymptotes of $$ f $$. 5.2 Write down the $$ y $$-intercept of $$ f $$. 5.3 Determine the $$ x $$-intercept of $$ f $$. 5.4 Sketch the graph of $$ f $$, showing ALL intercepts with the axes and any asymptotes. 5.5 Write down the domain of $$ h $$, if $$ h(x)=f(x-2) $$. (2)

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To find the equations of the asymptotes of the hyperbola $$ f(x)=\frac{3}{x-2}+1 $$, we need to identify the vertical and horizontal asymptotes.

1. **Vertical Asymptote:**
The vertical asymptote occurs when the denominator of the function becomes zero. In this case, the vertical asymptote is at $$ x = 2 $$.

2. **Horizontal Asymptote:**
To find the horizontal asymptote, we need to look at the behavior of the function as $$ x $$ approaches positive or negative infinity. As $$ x $$ approaches positive or negative infinity, the term $$ \frac{3}{x-2} $$ approaches 0. Therefore, the horizontal asymptote is at $$ y = 1 $$.

Now, let's move on to the other parts of the question:

5.2 **$$ y $$-intercept of $$ f $$:**
To find the $$ y $$-intercept, we set $$ x = 0 $$ in the function $$ f(x) $$ and solve for $$ y $$.

$$ f(0) = \frac{3}{0-2} + 1 = \frac{3}{-2} + 1 = -\frac{3}{2} + 1 = -\frac{1}{2} $$

So, the $$ y $$-intercept is at $$ (0, -\frac{1}{2}) $$.

5.3 **$$ x $$-intercept of $$ f $$:**
To find the $$ x $$-intercept, we set $$ y = 0 $$ in the function $$ f(x) $$ and solve for $$ x $$.

$$ 0 = \frac{3}{x-2} + 1 $$
$$ -1 = \frac{3}{x-2} $$
$$ -x + 2 = 3 $$
$$ -x = 1 $$
$$ x = -1 $$

So, the $$ x $$-intercept is at $$ (-1, 0) $$.

5.4 **Sketch the graph of $$ f $$:**
To sketch the graph of $$ f $$, we plot the intercepts and asymptotes on the coordinate plane.

- Vertical asymptote at $$ x = 2 $$
- Horizontal asymptote at $$ y = 1 $$
- $$ y $$-intercept at $$ (0, -\frac{1}{2}) $$
- $$ x $$-intercept at $$ (-1, 0) $$

5.5 **Domain of $$ h $$:**
Given that $$ h(x) = f(x-2) $$, the domain of $$ h $$ is the set of all real numbers except the values that make the denominator of $$ f(x) $$ equal to zero. Since the denominator of $$ f(x) $$ is $$ x-2 $$, the domain of $$ h $$ is all real numbers except $$ x = 2 $$.
1. Vertical asymptote: $$ x = 2 $$ 2. Horizontal asymptote: $$ y = 1 $$ 3. $$ y $$-intercept: $$ (0, -\frac{1}{2}) $$ 4. $$ x $$-intercept: $$ (-1, 0) $$ 5. Domain of $$ h $$: All real numbers except $$ x = 2 $$

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