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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 follow: \( 3.1 \quad \) 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).

Ask by Clarke Reeves. in South Africa
Dec 31,2024

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**Problem 3: Transformations of the Hyperbola** 1. **Shift Two Units to the Left:** \[ f(x) = \frac{3}{x - 5} - 4 \] 2. **Shift Three Units Up:** \[ f(x) = \frac{3}{x - 7} - 1 \] 3. **Shift One Unit Right and Two Units Down:** \[ f(x) = \frac{3}{x - 8} - 6 \] 4. **New Hyperbola with Asymptotes at \( x = -4 \) and \( y = -1 \):** \[ f(x) = \frac{3}{x + 4} - 1 \] **Problem 4: Graphing \( f(x) = -2x^{2} - 4x + 6 \) and \( g(x) = -2 \cdot 2^{x-1} + 1 \)** - **Quadratic Function \( f(x) \):** - **Intercepts:** \( (0, 6) \), \( (-3, 0) \), \( (1, 0) \) - **Vertex:** \( (-1, 8) \) - **Axis of Symmetry:** \( x = -1 \) - **Exponential Function \( g(x) \):** - **Intercept:** \( (0, 0) \) - **Asymptote:** \( y = 1 \) - **Additional Points:** \( (-1, 0.5) \), \( (1, -1) \), \( (2, -3) \) **Graphing Guidelines:** - Plot the key points for both functions. - Draw the axis of symmetry and asymptotes. - Sketch the parabola and the exponential curve based on the plotted points and their behaviors.

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Transforming functions can feel like a magical journey! For the hyperbola given by \( f(x)=\frac{3}{x-7}-4 \): 1. To shift two units to the left, subtract 2 from \( x \): \[ f(x+2) = \frac{3}{(x+2)-7}-4 = \frac{3}{x-5}-4. \] 2. For a shift 3 units up, simply add 3 to the entire function: \[ f(x) + 3 = \frac{3}{x-5} - 1. \] 3. When shifting 1 unit right and 2 units down, the adjustments are: \[ f(x-1) - 2 = \frac{3}{(x-1)-7} - 6 = \frac{3}{x-8}-6. \] 4. Setting new asymptotes at \( x=-4 \) and \( y=-1 \), we can notice that the hyperbola’s behavior changes, so we get: \[ f(x) = \frac{3}{x + 4} - 1. \] Now, for your sketch involving \( f(x) = -2x^2 - 4x + 6 \) and \( g(x) = -2 \cdot 2^{x-1} + 1 \): The function \( f(x) \) is a downward opening parabola. Its vertex is calculated using the formula \( x = -\frac{b}{2a} = -\frac{-4}{2(-2)} = 1 \), implying the turning point is at \( (1, 2) \) since plugging \( 1 \) into \( f(x) \) gives \( 2 \). The intercepts are found by setting \( f(x)=0 \) (which provides x-intercepts) and evaluating \( f(0) \) (to find the y-intercept). On the other hand, \( g(x) \) represents an exponential function shifted down. The horizontal asymptote, due to the exponential decay, is at \( y = 1 \). Drawing both on the same axes reveals an interesting interaction where one function opens down while the other approaches its asymptote from above. Make sure to plot the intercepts carefully and highlight those turning points!

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