Find the x-intercepts, axis of symmetry, range of f(x) = -x2 - x + 6, and the equation of g(x) passing through a point.

Step 1: Determine the x-intercepts of f. To find the x-intercepts, set f(x) = 0: -x^2 - x + 6 &= 0 \\ x^2 + x - 6 &= 0 \\ (x+3)(x-2) &= 0 This gives two x-intercepts: x = -3 or x = 2 The x-intercepts are (-3; 0) and (2; 0). Step 2: Write down the equation of the axis of symmetry of f. The axis of symmetry for a parabola f(x) = ax^2 + bx + c is given by the formula x = -(b)/(2a). For f(x) = -x^2 - x + 6, we have a = -1 and b = -1. x = -(-1)/(2(-1)) = -(-1)/(-2) = -(1)/(2) The equation of the axis of symmetry is x = -(1)/(2). Step 3: Determine the range of f. Since a = -1 < 0, the parabola opens downwards, meaning it has a maximum value. The maximum value occurs at the axis of symmetry, x = -(1)/(2). Substitute x = -(1)/(2) into f(x): f(-(1)/(2)) &= -(-(1)/(2))^2 - (-(1)/(2)) + 6 \\ &= -(1)/(4) + (1)/(2) + 6 \\ &= -(1)/(4) + (2)/(4) + (24)/(4) \\ &= (25)/(4) The maximum y-value is (25)/(4). Since the parabola opens downwards, the range is all y-values less than or equal to this maximum. The range of f is y (25)/(4). Step 4: Write down the equation of g in the form g(x) = mx + c. The line g(x) passes through point A(-2; 4). From the graph, D is the y-intercept of f(x). To find D, set x=0 in f(x): f(0) = -(0)^2 - (0) + 6 = 6 So, point D is (0; 6). Now, calculate the gradient m of the line passing through A(-2; 4) and D(0; 6): m = (y_2 - y_1)/(x_2 - x_1) = (6 - 4)/(0 - (-2)) = (2)/(2) = 1 Since D(0; 6) is the y-intercept, c=6. The equation of g is g(x) = x + 6. Step 5: Write down the average gradient between points A and D. The average gradient between points A and D is the slope of the line segment connecting them. Since A and D are points on the line g(x), this is simply the gradient of g(x). From Step 4, the gradient is m=1. The average gradient between A and D is 1. Step 6: Determine the equation of h, if h is the reflection of f about the x-axis and then translated 3 units to the right. Leave your answer in the form h(x) = a(x+p)^2 + q. First, write f(x) in the form f(x) = a(x+p)^2 + q. The turning point of f(x) is (-(1)/(2), (25)/(4)) and a=-1. f(x) = -(x - (-(1)/(2)))^2 + (25)/(4) = -(x + (1)/(2))^2 + (25)/(4) Reflection about the x-axis means y -y, so f(x) -f(x): -f(x) = - ( -(x + (1)/(2))^2 + (25)/(4) ) = (x + (1)/(2))^2 - (25)/(4) Translation 3 units to the right means replacing x with (x-3): h(x) &= ((x-3) + (1)/(2))^2 - (25)/(4) \\ &= (x - (6)/(2) + (1)/(2))^2 - (25)/(4) \\ &= (x - (5)/(2))^2 - (25)/(4) The equation of h is h(x) = (x - (5)/(2))^2 - (25)/(4). Step 7: Write down the values of x for which f(x) > 0. From Step 1, the x-intercepts of f(x) are x = -3 and x = 2. Since f(x) = -x^2 - x + 6 is a parabola opening downwards, f(x) > 0 for the x-values between its x-intercepts. The values of x for which f(x) > 0 are -3 < x < 2. Step 8: If f(p) = f(r) = 4, calculate the value of p-r if r < 0. Set f(x) = 4: -x^2 - x + 6 &= 4 \\ -x^2 - x + 2 &= 0 \\ x^2 + x - 2 &= 0 \\ (x+2)(x-1) &= 0 The solutions are x = -2 or x = 1. We are given f(p) = 4 and f(r) = 4, so p and r are these two x-values. We are also given that r < 0. Therefore, r = -2. This means p must be the other value