The function f defined by f(x) = (a)/(x+p) + q has range y != 2, axis of symmetry y = x + 1 (positive gradient), and passes through (0, -4). Find a, p, q.

QUESTION 6 The function f defined by f(x) = (a)/(x+p) + q has the following properties: • The range of f is y R, y ≠ 2 • The axis of symmetry with a positive gradient is y = x + 1 • The graph of f passes through (0; -4) 6.1 Write down the value of q. Step 1: Identify the horizontal asymptote from the range. For a function of the form f(x) = (a)/(x+p) + q, the horizontal asymptote is y=q. The range is given as y R, y ≠ 2. Therefore, q = 2. q = 2 6.2 Calculate the values of a and p. Step 1: Use the axis of symmetry to find p. The axes of symmetry for a hyperbola f(x) = (a)/(x+p) + q are y - q = ± (x + p). We know q=2, so the axes are y - 2 = ± (x + p). The axis of symmetry with a positive gradient is y - 2 = x + p. Rearranging this gives y = x + p + 2. We are given that this axis of symmetry is y = x + 1. Comparing the constant terms: p + 2 = 1 p = 1 - 2 p = -1 Step 2: Use the given point to find a. Now the function is f(x) = (a)/(x-1) + 2. The graph passes through the point (0; -4). Substitute these coordinates into the function: -4 = (a)/(0-1) + 2 -4 = (a)/(-1) + 2 -4 = -a + 2 -4 - 2 = -a -6 = -a 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 q=2, the horizontal asymptote is y=2. From p=-1, the vertical asymptote is x = -p = -(-1) = 1. Step 2: Calculate the intercepts. The function is f(x) = (6)/(x-1) + 2. Y-intercept (set x=0): This is given as (0, -4). f(0) = (6)/(0-1) + 2 = -6 + 2 = -4. So the y-intercept is (0, -4). X-intercept (set y=0): 0 = (6)/(x-1) + 2 -2 = (6)/(x-1) -2(x-1) = 6 -2x + 2 = 6 -2x = 4 x = -2 So the x-intercept is (-2, 0). Step 3: Sketch the graph. Draw the Cartesian plane. Draw the vertical asymptote x=1 as a dashed line. Draw the horizontal asymptote y=2 as a dashed line. Plot the x-intercept (-2, 0). Plot the y-intercept (0, -4). Since a=6 is positive, the branches of the hyperbola will be in the first and third quadrants relative to the asymptotes. Sketch the curve passing through the intercepts and approaching the asymptotes. For a point in the first quadrant relative to the asymptotes, let x=2: f(2) = (6)/(2-1) + 2 = 6 + 2 = 8. Plot (2, 8). `mermaid graph TD A[Start] --> BDraw Axes; B --> CDraw Vertical Asymptote x=1; C --> DDraw Horizontal Asymptote y=2; D --> EPlot X-intercept (-2, 0); E --> FPlot Y-intercept (0, -4); F --> GPlot additional point (2, 8); G --> HSketch Hyperbola Branch 1 (through (-2,0) and (0,-4)); H --> ISketch Hyperbola Branch 2 (through (2,8)); I --> J[End]; ` The graph should look like this: (Due to text-based limitations, I cannot draw the graph here. Please sketch it on paper following the steps above.) Send me the next one 📸