This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — Question 5 5.1 Write down the coordinates of B. Step 1: Identify B as the y-intercept. The y-intercept occurs when x=0. Step 2: Substitute x=0 into the function g(x) = a^x. g(0) = a^0 = 1 The coordinates of B are (0; 1). 5.2 Determine the value of a. Step 1: Use the given point A(-3; (27)/(8)) which lies on g(x) = a^x. Substitute the coordinates of A into the equation: (27)/(8) = a^-3 Step 2: Solve for a. (27)/(8) = (1)/(a^3) a^3 = (8)/(27) a = 3/(27) a = (2)/(3) The value of a is (2)/(3). 5.3 Determine the equation of h if h is the reflection of g in the y-axis. Step 1: Understand reflection in the y-axis. A reflection in the y-axis means replacing x with -x in the function's equation. Step 2: Apply the reflection to g(x) = ((2)/(3))^x. h(x) = g(-x) = ((2)/(3))^-x h(x) = ((3)/(2))^x The equation of h is h(x) = ((3)/(2))^x. 5.4 Write down the equation of h^-1(x), the inverse of h, in the form y = . Step 1: Start with the equation of h(x) and replace h(x) with y. y = ((3)/(2))^x Step 2: Swap x and y to find the inverse. x = ((3)/(2))^y Step 3: Solve for y by converting the exponential equation to logarithmic form. y = _(3)/(2) x The equation of h^-1(x) is y = _(3)/(2) x. 5.5 For which values of x will h^-1(x) 1? Step 1: Set up the inequality using the inverse function. _(3)/(2) x 1 Step 2: Convert the logarithmic inequality to an exponential inequality. Since the base (3)/(2) > 1, the inequality direction remains the same. x ((3)/(2))^1 x (3)/(2) Step 3: Consider the domain of the logarithmic function. For h^-1(x) = _(3)/(2) x, the domain requires x > 0. Combining the conditions, 0 < x (3)/(2). The values of x are 0 < x (3)/(2). 5.6 Write down the domain of h^-1(x-3). Step 1: Recall the domain of h^-1(x). The domain of h^-1(x) = _(3)/(2) x is x > 0. Step 2: Apply the domain restriction to the argument of h^-1(x-3). For h^-1(x-3), the expression inside the logarithm, (x-3), must be greater than 0. x-3 > 0 x > 3 The domain of h^-1(x-3) is x > 3. Question 6 The function f is defined by f(x) = (a)/(x+p) + q. 6.1 Write down the value of q. Step 1: Identify the relationship between the range and the horizontal asymptote. For a hyperbola f(x) = (a)/(x+p) + q, the horizontal asymptote is y=q. The range is y R, y q. Step 2: Use the given range. The range of f is y R, y 2. Therefore, q=2. The value of q is 2. 6.2 Calculate the values of a and p. Step 1: Use the horizontal asymptote to update the function. From 6.1, q=2, so f(x) = (a)/(x+p) + 2. Step 2: Use the axis of symmetry to find p. The vertical asymptote is x = -p. The point of intersection of the asymptotes is (-p; q), which is (-p; 2). The axes of symmetry pass through this point. The given axis of symmetry is y = x+1. Substitute (-p; 2) into y = x+1: 2 = -p + 1 p = 1 - 2 p = -1 Step 3: Use the point the graph passes through to find a. Now the function is f(x) = (a)/(x-1) + 2. The graph passes through (0; -4). Substitute this point into the function: -4 = (a)/(0-1) + 2 -4 = (a)/(-1) + 2 -4 = -a + 2 -a = -6 a = 6 The values are a=6 and p=-1. 6.3 Sketch a neat graph of this function. Your graph must include the intercepts with the axes and asymptotes if any. Step 1: Identify the asymptotes. From f(x) = (6)/(x-1) + 2: Vertical asymptote: x=1. Horizontal asymptote: y=2. Step 2: Determine the intercepts. Y-intercept (set x=0): f(0) = (6)/(0-1) + 2 = -6 + 2 = -4 The y-intercept is (0; -4). X-intercept (set f(x)=0): 0 = (6)/(x-1) + 2 -2 = (6)/(x-1) -2(x-1) = 6 -2x + 2 = 6 -2x = 4 x = -2 The x-intercept is (-2; 0). Step 3: Sketch the graph. Plot the vertical asymptote x=1 and the horizontal asymptote y=2. Plot the intercepts (0; -4) and (-2; 0). Draw the two branches of the hyperbola, approaching the asymptotes and passing through the intercepts. `mermaid graph TD A[Start] --> BPlot Asymptotes; B --> C[Vertical: x=1]; B --> D[Horizontal: y=2]; B --> EFind Intercepts; E --> F[Y-intercept: (0, -4)]; E --> G[X-intercept: (-2, 0)]; E --> HSketch Hyperbola Branches; H --> I[Pass through intercepts]; H --> J[Approach asymptotes]; J --> K[End]; style A fill:#fff,stroke:#333,stroke-width:2px style B fill:#fff,stroke:#333,stroke-width:2px style C fill:#fff,stroke:#333,stroke-width:2px style D fill:#fff,stroke:#333,stroke-width:2px style E fill:#fff,stroke:#333,stroke-width:2px style F fill:#fff,stroke:#333,stroke-width:2px style G fill:#fff,stroke:#333,stroke-width:2px style H fill:#fff,stroke:#333,stroke-width:2px style I fill:#fff,stroke:#333,stroke-width:2px style J fill:#fff,stroke:#333,stroke-width:2px style K fill:#fff,stroke:#333,stroke-width:2px ` The graph should look like this: (Due to text-based format, I cannot draw the graph directly. Please sketch it on graph paper using the information above.) Send me the next one 📸