The equation of a hyperbola is given by \( f(x)=\frac{x^{x-7}}{}-4 \). 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 \( 3.4 \quad \) The equation of the new hyperbola has new asymptotes at \( x=-4 \) and \( y=-1 \) 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).

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!