QUESTION 3 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 transformed as follows: \( 3.1 \quad \) Shift two units to the left \( 3.2 \quad \) Shift 3 units up \( 3.3 \quad \) Shift 1 unit right and 2 units down QUESTION 4 Sketch on the same set of axes the graphs of \( f(x)=-2 x^{2}-4 x+6 \) and \( g(x)=-2 \cdot 2^{x-1}+1 \). Clearly indicate all intercepts with the axes, turning point(s) and asymptote(s). [8]

### **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.