QUESTION 3
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
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).
[8]

Ask by Chan Schmidt. in South Africa

Jan 23,2025

Question

QUESTION 3
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
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).
[8]

Ask by Chan Schmidt. in South Africa

Jan 23,2025

UpStudy AI · Accepted Answer

### **QUESTION 3**

Given the hyperbola $$ f(x) = \frac{3}{x - 7} - 4 $$, we'll apply the specified transformations to find the new equations.

#### **3.1 Shift Two Units to the Left**

**Transformation Details:**
- **Horizontal Shift:** Shifting left by 2 units involves replacing $$ x $$ with $$ x + 2 $$.

**New Equation:**
$$
f_{\text{new}}(x) = \frac{3}{(x + 2) - 7} - 4 = \frac{3}{x - 5} - 4
$$

---

#### **3.2 Shift Three Units Up**

**Transformation Details:**
- **Vertical Shift:** Shifting up by 3 units involves adding 3 to the entire function.

**New Equation:**
$$
f_{\text{new}}(x) = \frac{3}{x - 7} - 4 + 3 = \frac{3}{x - 7} - 1
$$

---

#### **3.3 Shift One Unit Right and Two Units Down**

**Transformation Details:**
1. **Horizontal Shift:** Shifting right by 1 unit replaces $$ x $$ with $$ x - 1 $$.
2. **Vertical Shift:** Shifting down by 2 units involves subtracting 2 from the entire function.

**New Equation:**
$$
f_{\text{new}}(x) = \frac{3}{(x - 1) - 7} - 4 - 2 = \frac{3}{x - 8} - 6
$$

---

### **QUESTION 4**

We are to sketch the graphs of the functions $$ f(x) = -2x^2 - 4x + 6 $$ and $$ g(x) = -2 \cdot 2^{x-1} + 1 $$ on the same set of axes, indicating all intercepts, turning points, and asymptotes.

#### **1. Graph of $$ f(x) = -2x^2 - 4x + 6 $$**

**Type of Function:** Quadratic (Parabola) opening downward.

**Key Features:**

- **Vertex (Turning Point):**
  $$
  x = -\frac{b}{2a} = -\frac{-4}{2(-2)} = -1
  $$
  $$
  f(-1) = -2(-1)^2 - 4(-1) + 6 = -2 + 4 + 6 = 8
  $$
  **Vertex:** $$ (-1, 8) $$

- **X-Intercepts:**
  Solve $$ -2x^2 - 4x + 6 = 0 $$:
  $$
  x^2 + 2x - 3 = 0 \quad \Rightarrow \quad (x + 3)(x - 1) = 0
  $$
  **X-Intercepts:** $$ x = -3 $$ and $$ x = 1 $$ $$ \Rightarrow (-3, 0) $$ and $$ (1, 0) $$

- **Y-Intercept:**
  $$
  f(0) = -2(0)^2 - 4(0) + 6 = 6
  $$
  **Y-Intercept:** $$ (0, 6) $$

**Graph Characteristics:**
- **Opens downward** due to the negative coefficient of $$ x^2 $$.
- **Symmetry axis** is the vertical line $$ x = -1 $$.

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

**Type of Function:** Exponential function.

**Key Features:**

- **Horizontal Asymptote:**
  $$
  y = 1
  $$

- **Y-Intercept:**
  $$
  g(0) = -2 \cdot 2^{-1} + 1 = -1 + 1 = 0
  $$
  **Y-Intercept:** $$ (0, 0) $$

- **X-Intercept:**
  Set $$ g(x) = 0 $$:
  $$
  -2 \cdot 2^{x-1} + 1 = 0 \quad \Rightarrow \quad 2^{x-1} = \frac{1}{2} \quad \Rightarrow \quad x - 1 = -1 \quad \Rightarrow \quad x = 0
  $$
  **X-Intercept:** $$ (0, 0) $$

- **Behavior:**
  - Since the coefficient of the exponential term is negative ($$-2$$), the function **decreases** as $$ x $$ increases.
  - **No turning points**; exponential functions are smooth and continuously increasing or decreasing.

**Graph Characteristics:**
- **Approaches the horizontal asymptote** $$ y = 1 $$ as $$ x \to \infty $$ and $$ y = -\infty $$ as $$ x \to -\infty $$.
- **Passes through the origin** $$ (0, 0) $$.

#### **Combined Graph Overview**

When sketching both functions on the same set of axes:

- **Quadratic $$ f(x) $$:**
  - Plot the vertex at $$ (-1, 8) $$.
  - Plot the x-intercepts at $$ (-3, 0) $$ and $$ (1, 0) $$.
  - Plot the y-intercept at $$ (0, 6) $$.
  - Draw a downward-opening parabola symmetric about $$ x = -1 $$.

- **Exponential $$ g(x) $$:**
  - Plot the point $$ (0, 0) $$.
  - Draw the horizontal asymptote at $$ y = 1 $$.
  - Sketch the decreasing exponential curve approaching $$ y = 1 $$ as $$ x $$ increases and descending without bound as $$ x $$ decreases.

**Intercepts and Asymptotes:**

- **$$ f(x) $$:**
  - **X-Intercepts:** $$ (-3, 0) $$, $$ (1, 0) $$
  - **Y-Intercept:** $$ (0, 6) $$
  - **Vertex:** $$ (-1, 8) $$

- **$$ g(x) $$:**
  - **X-Intercept & Y-Intercept:** $$ (0, 0) $$
  - **Horizontal Asymptote:** $$ y = 1 $$

**Turning Points:**

- **$$ f(x) $$:** One turning point at the vertex $$ (-1, 8) $$.
- **$$ g(x) $$:** No turning points.

---

### **Summary**

- **Question 3:** The equations after the specified transformations are:
  1. **3.1 Shift Left 2 Units:** $$ f(x) = \frac{3}{x - 5} - 4 $$
  2. **3.2 Shift Up 3 Units:** $$ f(x) = \frac{3}{x - 7} - 1 $$
  3. **3.3 Shift Right 1 Unit and Down 2 Units:** $$ f(x) = \frac{3}{x - 8} - 6 $$

- **Question 4:** 
  - **For $$ f(x) $$:** A downward-opening parabola with vertex at $$ (-1, 8) $$, x-intercepts at $$ (-3, 0) $$ and $$ (1, 0) $$, and y-intercept at $$ (0, 6) $$.
  - **For $$ g(x) $$:** A decreasing exponential curve passing through $$ (0, 0) $$, with a horizontal asymptote at $$ y = 1 $$, and no turning points.
**Question 3:** After the transformations: 1. **Shift 2 units to the left:** $$ f(x) = \frac{3}{x - 5} - 4 $$ 2. **Shift 3 units up:** $$ f(x) = \frac{3}{x - 7} - 1 $$ 3. **Shift 1 unit right and 2 units down:** $$ f(x) = \frac{3}{x - 8} - 6 $$ **Question 4:** - **For $$ f(x) = -2x^2 - 4x + 6 $$:** - **Vertex:** $$ (-1, 8) $$ - **X-Intercepts:** $$ (-3, 0) $$ and $$ (1, 0) $$ - **Y-Intercept:** $$ (0, 6) $$ - **Graph:** Downward-opening parabola symmetric about $$ x = -1 $$ - **For $$ g(x) = -2 \cdot 2^{x-1} + 1 $$:** - **Y-Intercept:** $$ (0, 0) $$ - **Horizontal Asymptote:** $$ y = 1 $$ - **Graph:** Decreasing exponential curve approaching $$ y = 1 $$ as $$ x $$ increases

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