The graph of the function f(x) = racax+p + q has a vertical asymptote at x = -1, a horizontal asymptote at y = 1, and passes through the point P(-2; 4). Determine the values of a, p, and q.

Here's the solution to Question 1: 1.1 Determine the values of a, p, and q. Step 1: Identify p from the vertical asymptote. The general form of a hyperbola is f(x) = (a)/(x+p) + q. The vertical asymptote is given by x+p = 0, so x = -p. From the problem, the vertical asymptote is x = -1. Therefore, -p = -1 p = 1. Step 2: Identify q from the horizontal asymptote. The horizontal asymptote is y = q. From the problem, the horizontal asymptote is y = 1. Therefore, q = 1. Step 3: Use the given point P(-2; 4) to find a. Substitute p=1 and q=1 into the function: f(x) = (a)/(x+1) + 1 The point P(-2; 4) lies on the graph, so substitute x=-2 and f(x)=4: 4 = (a)/(-2+1) + 1 4 = (a)/(-1) + 1 4 = -a + 1 4 - 1 = -a 3 = -a a = -3 The values are: a = -3, p = 1, q = 1 1.2 Calculate the coordinates of T, the x-intercept of f. Step 1: Set f(x) = 0 to find the x-intercept. Using the function f(x) = (-3)/(x+1) + 1: 0 = (-3)/(x+1) + 1 Step 2: Solve for x. -1 = (-3)/(x+1) Multiply both sides by (x+1): -(x+1) = -3 x+1 = 3 x = 3 - 1 x = 2 The x-intercept is at x=2, and the y-coordinate is 0. The coordinates of T are: (2; 0) 1.3 If the graph of f is symmetrical with respect to the line y = x+c, determine the value of c. Step 1: Identify the point of intersection of the asymptotes. The asymptotes are x = -1 and y = 1. Their intersection point is (-1; 1). For a hyperbola, the lines of symmetry pass through the point of intersection of its asymptotes. Step 2: Use the point of intersection of asymptotes to find c. The given line of symmetry is y = x+c. Since this line passes through (-1; 1), substitute x=-1 and y=1 into the equation: 1 = -1 + c c = 1 + 1 c = 2 The value of c is: 2