The equation of a hyperbola is given by .
Write down the equation of the new function that is formed when is transformed as follows:
Shift two units to the left
Shift 3 units up
Shift 1 unit right and 2 units down
The equation of the new hyperbola has new asymptotes at and
QUESTION 4
Sketch on the same set of axes the graphs of and
Clearly indicate all intercepts with the axes, turning point(s) and asymptote(s).

Ask by Deleon Cole. in South Africa

Jan 23,2025

Question

The equation of a hyperbola is given by .
Write down the equation of the new function that is formed when is transformed as follows:
Shift two units to the left
Shift 3 units up
Shift 1 unit right and 2 units down
The equation of the new hyperbola has new asymptotes at and
QUESTION 4
Sketch on the same set of axes the graphs of and
Clearly indicate all intercepts with the axes, turning point(s) and asymptote(s).

Ask by Deleon Cole. in South Africa

Jan 23,2025

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Part 3: Transforming the Function

The original function is given by:

$$
f(x) = \frac{x^{x-7}}{} - 4
$$

However, it seems there is a missing part in the function definition. Assuming the function is meant to be $$ f(x) = \frac{x^{x-7}}{1} - 4 $$ (which simplifies to $$ f(x) = x^{x-7} - 4 $$), we will proceed with the transformations.

#### 3.1: Shift two units to the left

To shift a function $$ f(x) $$ to the left by $$ h $$ units, we replace $$ x $$ with $$ x + h $$. Here, $$ h = 2 $$:

$$
f_1(x) = f(x + 2) = (x + 2)^{(x + 2) - 7} - 4 = (x + 2)^{x - 5} - 4
$$

#### 3.2: Shift 3 units up

To shift a function $$ f(x) $$ up by $$ k $$ units, we add $$ k $$ to the function. Here, $$ k = 3 $$:

$$
f_2(x) = f_1(x) + 3 = (x + 2)^{x - 5} - 4 + 3 = (x + 2)^{x - 5} - 1
$$

#### 3.3: Shift 1 unit right and 2 units down

First, we shift right by 1 unit:

$$
f_3(x) = f_2(x - 1) = (x - 1 + 2)^{(x - 1) - 5} - 1 = (x + 1)^{x - 6} - 1
$$

Now, we shift down by 2 units:

$$
f_4(x) = f_3(x) - 2 = (x + 1)^{x - 6} - 1 - 2 = (x + 1)^{x - 6} - 3
$$

#### 3.4: New asymptotes at $$ x = -4 $$ and $$ y = -1 $$

To achieve new asymptotes, we need to adjust the function. The vertical asymptote at $$ x = -4 $$ suggests a transformation of the form $$ f(x) = (x + 4)^{x - k} - 1 $$ for some $$ k $$. The horizontal asymptote at $$ y = -1 $$ suggests that as $$ x \to \infty $$, the function approaches -1.

Thus, the new function can be expressed as:

$$
f_5(x) = (x + 4)^{x - 6} - 1
$$

### Part 4: Sketching the Graphs

Now, we need to sketch the graphs of the two functions:

1. **Function $$ f(x) = -2x^2 - 4x + 6 $$**

- This is a quadratic function. To find the intercepts:
     - **Y-intercept**: Set $$ x = 0 $$:
       $$
       f(0) = -2(0)^2 - 4(0) + 6 = 6 \quad \text{(Y-intercept at (0, 6))}
       $$
     - **X-intercepts**: Set $$ f(x) = 0 $$:
       $$
       -2x^2 - 4x + 6 = 0 \implies x^2 + 2x - 3 = 0
       $$
       Using the quadratic formula:
       $$
       x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 + 24}}{2} = \frac{-2 \pm 2\sqrt{7}}{2} = -1 \pm \sqrt{7}
       $$
       Thus, the X-intercepts are $$ x = -1 + \sqrt{7} $$ and $$ x = -1 - \sqrt{7} $$.

- **Turning Point**: The vertex can be found using $$ x = -\frac{b}{2a} = -\frac{-4}{2(-2)} = 1 $$:
     $$
     f(1) = -2(1)^2 - 4(1) + 6 = 0 \quad \text{(Turning point at (1, 0))}
     $$

2. **Function $$ g(x) = -2 \cdot 2^{x-1} + 1 $$**

- **Y-intercept**: Set $$ x = 0 $$:
     $$
     g(0) = -2 \cdot 2^{-1} + 1 = -1 + 1 = 0 \quad \text{(Y-intercept at (0, 0))}
     $$
   - **Asymptote**: As $$ x \to -\infty $$, $$ g(x) \to 1 $$ (horizontal asymptote at $$ y = 1 $$).

### Summary of Key Points

- **Function $$ f(x) $$**:
  - Y-intercept: (0, 6)
  - X-intercepts: $$ x = -1 + \sqrt{7}, -1 - \sqrt{7} $$
  - Turning point: (1, 0)

- **Function $$ g(x) $$**:
  - Y-intercept: (0, 0)
  - Horizontal asymptote: $$ y = 1 $$

### Sketching the Graphs

To sketch the graphs, plot the intercepts, turning points, and asymptotes on the same set of axes. The parabola will open downwards, while the exponential function will approach its horizontal asymptote as $$ x $$ increases.

If you need a visual representation, please let me know!
After transforming the original function $$ f(x) = x^{x-7} - 4 $$ through shifting two units to the left, three units up, one unit right, and two units down, the new function becomes $$ f(x) = (x + 1)^{x - 6} - 3 $$. For the graphs of $$ f(x) = -2x^2 - 4x + 6 $$ and $$ g(x) = -2 \cdot 2^{x-1} + 1 $$: - **Function $$ f(x) $$**: - Y-intercept at (0, 6) - X-intercepts at $$ x = -1 + \sqrt{7} $$ and $$ x = -1 - \sqrt{7} $$ - Turning point at (1, 0) - **Function $$ g(x) $$**: - Y-intercept at (0, 0) - Horizontal asymptote at $$ y = 1 $$ These key points should be plotted on the same set of axes to sketch the graphs accurately.

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