This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Calculate the coordinates of B, the turning point of f. The function is given by f(x) = -x^2 + 4x + 5. For a parabola in the form ax^2 + bx + c, the x-coordinate of the turning point is given by x = -(b)/(2a). Here, a = -1 and b = 4. x_B = -(4)/(2(-1)) = -(4)/(-2) = 2 Substitute x_B = 2 into f(x) to find the y-coordinate: y_B = f(2) = -(2)^2 + 4(2) + 5 = -4 + 8 + 5 = 9 The coordinates of B are (2, 9). B(2, 9) Step 2: Show that the equation of the line through A and C is given by g(x) = 2x + 2. The point C is given as (3, 8). To find point A, we need to find the x-intercepts of f(x) by setting f(x) = 0: -x^2 + 4x + 5 = 0 Multiply by -1: x^2 - 4x - 5 = 0 Factor the quadratic equation: (x - 5)(x + 1) = 0 The x-intercepts are x = 5 and x = -1. From the diagram, A is the negative x-intercept, so A is (-1, 0). Now, calculate the slope m of the line passing through A(-1, 0) and C(3, 8): m = (y_2 - y_1)/(x_2 - x_1) = (8 - 0)/(3 - (-1)) = (8)/(4) = 2 Using the point-slope form y - y_1 = m(x - x_1) with point A(-1, 0): y - 0 = 2(x - (-1)) y = 2(x + 1) y = 2x + 2 Thus, the equation of the line is g(x) = 2x + 2. Step 3: Calculate the maximum length of EH for f > g. The length of EH is the vertical distance between f(x) and g(x), which is D(x) = f(x) - g(x). D(x) = (-x^2 + 4x + 5) - (2x + 2) D(x) = -x^2 + 4x + 5 - 2x - 2 D(x) = -x^2 + 2x + 3 This is a downward-opening parabola, so its maximum value occurs at its turning point. The x-coordinate of the turning point is x = -(b)/(2a). Here, a = -1 and b = 2. x = -(2)/(2(-1)) = -(2)/(-2) = 1 Substitute x = 1 into D(x) to find the maximum length: D(1) = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4 The maximum length of EH is 4 units. 4 units Step 4: Determine the value of m such that g is a tangent to k. The function k(x) is given as k(x) = f(x+m) = -x^2 - 2mx - m^2 + 4x + 4m + 5. We can rewrite k(x) by grouping terms: k(x) = -x^2 + (4 - 2m)x + (-m^2 + 4m + 5) For g(x) to be a tangent to k(x), there must be exactly one point of intersection between the two functions. Set k(x) = g(x): -x^2 + (4 - 2m)x + (-m^2 + 4m + 5) = 2x + 2 Rearrange the equation to form a quadratic equation Ax^2 + Bx + C = 0: -x^2 + (4 - 2m - 2)x + (-m^2 + 4m + 5 - 2) = 0 -x^2 + (2 - 2m)x + (-m^2 + 4m + 3) = 0 Multiply by -1 to make the leading coefficient positive: x^2 - (2 - 2m)x - (-m^2 + 4m + 3) = 0 x^2 + (2m - 2)x + (m^2 - 4m - 3) = 0 For a tangent, the discriminant ( = B^2 - 4AC) of this quadratic equation must be zero. Here, A = 1, B = (2m - 2), and C = (m^2 - 4m - 3). (2m - 2)^2 - 4(1)(m^2 - 4m - 3) = 0 Expand the terms: (4m^2 - 8m + 4) - (4m^2 - 16m - 12) = 0 4m^2 - 8m + 4 - 4m^2 + 16m + 12 = 0 Combine like terms: (-8m + 16m) + (4 + 12) = 0 8m + 16 = 0 8m = -16 m = -2 The value of m is -2. m = -2 Send me the next one 📸