The equation of a hyperbola is given by $$ f(x)=\frac{3}{x-7}-4 $$ Write down the equation of the new function that is formed when $$ f $$ is translormed as follows: 3.1 $$ \begin{array}{ll} ext { Shifi two units to the left } & ext { Shift } 3 ext { units up } \ 3.2 & ext { Shift } 1 ext { unit right and } 2 ext { units down } \ 3.4 & ext { The equation of the new hyperbola has new asymptotes at } x=-4 ext { and } y=-1 \ ext { asymptote }\end{array} $$

Question

The equation of a hyperbola is given by $$ f(x)=\frac{3}{x-7}-4 $$ Write down the equation of the new function that is formed when $$ f $$ is translormed as follows: 3.1 $$ \begin{array}{ll}	ext { Shifi two units to the left } & 	ext { Shift } 3 	ext { units up } \ 3.2 & 	ext { Shift } 1 	ext { unit right and } 2 	ext { units down } \ 3.4 & 	ext { The equation of the new hyperbola has new asymptotes at } x=-4 	ext { and } y=-1 \ 	ext { asymptote }\end{array} $$

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Let's address each transformation step by step for the hyperbola given by $$ f(x) = \frac{3}{x - 7} - 4 $$.

### 3.1 Shift Two Units to the Left and Three Units Up

1. **Shift Two Units to the Left:**
   - Replace $$ x $$ with $$ x + 2 $$ in the function.
   - New expression: $$ \frac{3}{(x + 2) - 7} - 4 = \frac{3}{x - 5} - 4 $$

2. **Shift Three Units Up:**
   - Add 3 to the entire function.
   - Final equation: $$ f_{\text{new}}(x) = \frac{3}{x - 5} - 1 $$

### 3.2 Shift One Unit to the Right and Two Units Down

1. **Shift One Unit to the Right:**
   - Replace $$ x $$ with $$ x - 1 $$ in the function.
   - New expression: $$ \frac{3}{(x - 1) - 7} - 4 = \frac{3}{x - 8} - 4 $$

2. **Shift Two Units Down:**
   - Subtract 2 from the entire function.
   - Final equation: $$ f_{\text{new}}(x) = \frac{3}{x - 8} - 6 $$

### 3.4 New Hyperbola with Asymptotes at $$ x = -4 $$ and $$ y = -1 $$

The general form of a hyperbola is:
$$ f(x) = \frac{A}{x - h} + k $$
where:
- $$ h $$ determines the vertical asymptote at $$ x = h $$,
- $$ k $$ determines the horizontal asymptote at $$ y = k $$,

Given the asymptotes $$ x = -4 $$ and $$ y = -1 $$:
- $$ h = -4 $$
- $$ k = -1 $$
- Assuming the same vertical stretch as the original function ($$ A = 3 $$),

The new equation is:
$$ f_{\text{new}}(x) = \frac{3}{x + 4} - 1 $$

### Summary of Answers

1. **Problem 3.1 Answer:**
   $$
   f_{\text{new}}(x) = \frac{3}{x - 5} - 1
   $$

2. **Problem 3.2 Answer:**
   $$
   f_{\text{new}}(x) = \frac{3}{x - 8} - 6
   $$

3. **Problem 3.4 Answer:**
   $$
   f_{\text{new}}(x) = \frac{3}{x + 4} - 1
   $$
The new equations after the transformations are: 1. **Shift two units to the left and three units up:** $$ f_{\text{new}}(x) = \frac{3}{x - 5} - 1 $$ 2. **Shift one unit to the right and two units down:** $$ f_{\text{new}}(x) = \frac{3}{x - 8} - 6 $$ 3. **New hyperbola with asymptotes at $$ x = -4 $$ and $$ y = -1 $$:** $$ f_{\text{new}}(x) = \frac{3}{x + 4} - 1 $$

The equation of a hyperbola is given by
Write down the equation of the new function that is formed when is translormed as follows:
3.1

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The equation of a hyperbola is given by Write down the equation of the new function that is formed when is translormed as follows: 3.1