The equation of the hyperbola is given as (x+2)(y+3) = t.

Here are the solutions to Question 1: 1.1.1 The equation of the hyperbola is given as (x+2)(y+3) = t. The vertical asymptote is found by setting the factor involving x to zero: x+2 = 0 x = -2 The horizontal asymptote is found by setting the factor involving y to zero: y+3 = 0 y = -3 The equations of the asymptotes of g are x = -2 and y = -3. 1.1.2 1.1.2.1 Determine the value of t. We are given that g(0) = -(5)/(2). This means when x=0, y=-(5)/(2). Substitute these values into the equation (x+2)(y+3) = t: (0+2)(-(5)/(2)+3) = t (2)(-(5)/(2)+(6)/(2)) = t (2)((1)/(2)) = t t = 1 1.1.2.2 Determine the values of p and q. The function f(x) = (x+p)^2 + q is a parabola. Its axis of symmetry is x = -p. The problem states that one of the asymptotes of g is an axis of symmetry for f. From 1.1.1, the asymptotes of g are x=-2 and y=-3. Since the axis of symmetry for a parabola is a vertical line, it must be x=-2. Therefore, -p = -2 p = 2. The turning point of the parabola f is (-p, q). Since p=2, the x-coordinate of the turning point is -2. From the graph, the turning point of f is at (-2, -1). Thus, q = -1. The values are p = 2 and q = -1. 1.2 Write g in the form y = (a)/(x+p) + q. From 1.1.2.1, we found t=1. So the equation for g is (x+2)(y+3) = 1. Divide by (x+2): y+3 = (1)/(x+2) Subtract 3 from both sides: y = (1)/(x+2) - 3 So, g in the required form is y = (1)/(x+2) - 3. 1.3 If h(x) = x-1 is the line of symmetry to g, determine the coordinates of the points of intersection h and g. To find the points of intersection, we set the equations for h(x) and g(x) equal to each other. Equation for h: y = x-1 Equation for g: (x+2)(y+3) = 1 (using t=1 from 1.1.2.1) Substitute y = x-1 into the equation for g: (x+2)((x-1)+3) = 1 (x+2)(x+2) = 1 (x+2)^2 = 1 Take the square root of both sides: x+2 = ± sqrt(1) x+2 = ± 1 Case 1: x+2 = 1 x = 1-2 x = -1 Substitute x=-1 into y=x-1: y = -1-1 y = -2 Point 1: (-1, -2) Case 2: x+2 = -1 x = -1-2 x = -3 Substitute x=-3 into y=x-1: y = -3-1 y = -4 Point 2: (-3, -4) The coordinates of the points of intersection are (-1, -2) and (-3, -4). 1.4 If k = x^2+4x+3. Determine the values of k if its roots are: We consider the quadratic equation x^2+4x+3-k=0. The discriminant is = B^2 - 4AC. Here, A=1, B=4, C=3-k. = (4)^2 - 4(1)(3-k) = 16 - 12 + 4k = 4 + 4k 1.4.1 non-real roots For non-real roots, the discriminant must be less than zero ( < 0). 4 + 4k < 0 4k < -4 k < -1 1.4.2 negative and unequal roots For unequal roots, the discriminant must be greater than zero ( > 0). 4 + 4k > 0 4k > -4 k > -1 For negative roots, two conditions must be met: 1. The sum of the roots (x_1+x_2 = -(B)/(A)) must be negative. x_1+x_2 = -(4)/(1) = -4 This condition is satisfied as -4 < 0. 2. The product of the roots (x_1x_2 = (C)/(A)) must be positive. x_1x_2 = (3-k)/(1) = 3-k 3-k > 0 -k > -3 k < 3 Combining the conditions k > -1 and k < 3, we get: -1 < k < 3 1.5 1.5.1 domain of g The function g(x) is a hyperbola with a vertical asymptote at x=-2. The function is undefined at this value. The domain of g is x R, x ≠ -2. 1.5.2 range of f The function f(x) = (x+p)^2 + q is a parabola. From 1.1.2.2, we found p=2 and q=-1. So, f(x) = (x+2)^2 - 1. This is a parabola that opens upwards (since the coefficient of (x+2)^2 is positive). Its minimum value occurs at the turning point (-2, -1). The minimum y-value is -1. The range of f is y -1. That's 2 down. 3 left today — send the next one.