Given the function f(x) = x2 - 2x - 3, determine its range, x-intercepts (D and E), and the equation of line g passing through P and E.

Here are the solutions to the problems: 6.1 The function f(x) = x^2 - 2x - 3 is a parabola opening upwards, as the coefficient of x^2 is positive. The turning point is given as T(1; -4). The range of f is all possible y-values. Since the parabola opens upwards, the minimum y-value is the y-coordinate of the turning point. The range of f is y -4. y -4 6.2 D and E are the x-intercepts of f(x). To find the x-intercepts, we set f(x) = 0. x^2 - 2x - 3 = 0 Step 1: Factor the quadratic equation. (x-3)(x+1) = 0 Step 2: Solve for x. x-3=0 or x+1=0 x=3 or x=-1 Step 3: Assign the coordinates based on the graph. From the graph, D is the negative x-intercept and E is the positive x-intercept. The coordinates are D(-1; 0) and E(3; 0). D(-1; 0), E(3; 0) 6.3 The equation of g is a straight line g(x) = mx+c. The graphs of f and g intersect at P and E. Step 1: Find the coordinates of P. P is the y-intercept of f(x). To find P, set x=0 in f(x). f(0) = (0)^2 - 2(0) - 3 = -3 So, P is (0; -3). Step 2: Use the coordinates of P and E to find the equation of g(x). The line g(x) passes through P(0; -3) and E(3; 0). The y-intercept c is the y-coordinate of P, so c = -3. Step 3: Calculate the slope m. m = (y_2 - y_1)/(x_2 - x_1) = (0 - (-3))/(3 - 0) = (3)/(3) = 1 Step 4: Write the equation of g(x). g(x) = 1x - 3 g(x) = x-3 6.4 We need to find the values of x for which f(x) - g(x) > 0, which means f(x) > g(x). This inequality holds when the graph of f(x) is above the graph of g(x). Step 1: Identify the intersection points of f(x) and g(x). The graphs intersect at P(0; -3) and E(3; 0). These are the points where f(x) = g(x). Step 2: Observe the graph to determine where f(x) is above g(x). From the graph, f(x) is above g(x) for x-values to the left of P and to the right of E. So, x < 0 or x > 3. x < 0 or x > 3 6.5 We need to determine the maximum vertical distance between h(x) and g(x) for x [-2; 3], where h(x) = -f(x). Step 1: Find the equation for h(x). h(x) = -f(x) = -(x^2 - 2x - 3) = -x^2 + 2x + 3 Step 2: Define the vertical distance function D(x). The vertical distance is D(x) = |h(x) - g(x)|. D(x) = |(-x^2 + 2x + 3) - (x-3)| D(x) = |-x^2 + 2x + 3 - x + 3| D(x) = |-x^2 + x + 6| Step 3: Find the maximum value of the expression inside the absolute value, k(x) = -x^2 + x + 6, on the interval x [-2; 3]. This is a downward-opening parabola. Its maximum value occurs at its vertex. The x-coordinate of the vertex is x = -(b)/(2a) = -(1)/(2(-1)) = (1)/(2). The y-coordinate of the vertex is k((1)/(2)) = -((1)/(2))^2 + ((1)/(2)) + 6 = -(1)/(4) + (2)/(4) + (24)/(4) = (25)/(4) = 6.25. Since x=(1)/(2) is within the interval [-2; 3], the maximum value of k(x) is 6.25. Step 4: Check the values at the endpoints of the interval. At x=-2: k(-2) = -(-2)^2 + (-2) + 6 = -4 - 2 + 6 = 0. At x=3: k(3) = -(3)^2 + (3) + 6 = -9 + 3 + 6 = 0. Step 5: Determine the maximum vertical distance. The maximum value of k(x) is 6.25, and since k(x) is positive on the interval (-2, 3), D(x) = k(x). The maximum vertical distance is 6.25. 6.25 6.6 Given k(x) = g(x) - n. We need to determine n if k(x) is a tangent to f(x). Step 1: Write the equation for k(x). k(x) = (x-3) - n = x - 3 - n Step 2: Find the derivatives of f(x) and k(x). f(x) = x^2 - 2x - 3 f'(x) = 2x - 2 k(x) = x - 3 - n k'(x) = 1 Step 3: For k(x) to be tangent to f(x), their slopes must be equal at the point of tangency. f'(x) = k'(x) 2x - 2 = 1 2x = 3 x = (3)/(2) This is the x-coordinate of the point of tangency. Step 4: At the point of tangency, the y-values of f(x) and k(x) must be equal. f((3)/(2)) = k((3)/(2)) Substitute $x = \