A function g has equation (x+2)(y+3)=t. The function f(x)=(x+p)2+q has an axis of symmetry which is one of the asymptotes of g. If g(0) = -5/2, determine the values of t, p, q.

Here's the continuation of the solution: 1.1.1 The equations of the asymptotes of g are derived from the given equation (x+2)(y+3) = t. The vertical asymptote occurs when the term with x is zero. The horizontal asymptote occurs when the term with y is zero. Step 1: Identify the asymptotes from the equation. For g(x): (x+2)(y+3) = t: The vertical asymptote is x+2=0 x=-2. The horizontal asymptote is y+3=0 y=-3. The equations of the asymptotes are x=-2 and y=-3. 1.1.2 Determine the values of: 1.1.2.1 t We are given that g(0) = -(5)/(2). This means the point (0, -(5)/(2)) lies on the graph of g. Step 1: Substitute the point (0, -(5)/(2)) into the equation for g(x). (x+2)(y+3) = t (0+2)(-(5)/(2)+3) = t Step 2: Simplify the expression to find t. 2(-(5)/(2)+(6)/(2)) = t 2((1)/(2)) = t t = 1 1.1.2.2 p and q The function f(x) = (x+p)^2 + q is a parabola. The problem states that one of the asymptotes of g is an axis of symmetry for f. Step 1: Identify the axis of symmetry for f and relate it to the asymptotes of g. The axis of symmetry for f(x) = (x+p)^2 + q is the vertical line x = -p. 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. Step 2: Determine the value of q using the graph. The vertex of the parabola f(x) = (x+p)^2 + q is at (-p, q). With p=2, the vertex is at (-2, q). From the sketch, the parabola f touches the x-axis at x=-2, which means its vertex is at (-2, 0). Therefore, q=0. The values are p=2 and q=0. That's 2 down. 3 left today — send the next one.