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.
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Here are the solutions for questions 45 through 50. Question 45: Step 1: Determine the slope of the tangent at the given point. From question 44, the derivative of y = (x)/(x-2) is (dy)/(dx) = (-2)/((x-2)^2). At the point (4, 2), the slope of the tangent m_tangent is: m_tangent = (-2)/((4-2)^2) = (-2)/(2^2) = (-2)/(4) = -(1)/(2) Step 2: Calculate the slope of the normal. The normal line is perpendicular to the tangent line, so m_normal = -(1)/(m_tangent). m_normal = -(1)/(-1)2 = 2 Step 3: Use the point-slope form to find the equation of the normal. Using the point (x_1, y_1) = (4, 2) and the slope m_normal = 2: y - 2 = 2(x - 4) y - 2 = 2x - 8 2x - y - 6 = 0 The equation of the normal is: 2x - y - 6 = 0 Question 46: Step 1: Find the first derivative (dy)/(dx) using the quotient rule. Given y = (x^2)/(x-1). (dy)/(dx) = ((2x)(x-1) - (x^2)(1))/((x-1)^2) = (2x^2 - 2x - x^2)/((x-1)^2) = (x^2 - 2x)/((x-1)^2) Step 2: Evaluate the derivative at the point (2, 4) to find the slope of the tangent. Substitute x=2 into the derivative: m = ((2)^2 - 2(2))/((2-1)^2) = (4 - 4)/((1)^2) = (0)/(1) = 0 Step 3: Use the point-slope form to find the equation of the tangent. Using the point (x_1, y_1) = (2, 4) and the slope m = 0: y - 4 = 0(x - 2) y - 4 = 0 y = 4 The equation of the tangent is: y = 4 Question 47: Step 1: Determine the slope of the tangent at the given point. From question 46, the slope of the tangent m_tangent at (2, 4) is 0. Step 2: Calculate the slope of the normal. Since the tangent is a horizontal line (y=4), the normal line will be a vertical line with an undefined slope. Step 3: Find the equation of the normal. A vertical line passing through the point (2, 4) has the equation x = x_1. x = 2 The equation of the normal is: x = 2 Question 48: Step 1: Find the first derivative (dy)/(dx). Given y = (x^2+1)/(x). Rewrite as y = x + x^-1. Differentiate y with respect to x: (dy)/(dx) = 1 - x^-2 = 1 - (1)/(x^2) Step 2: Evaluate the derivative at the point (1, 2) to find the slope of the tangent. Substitute x=1 into the derivative: m = 1 - (1)/((1)^2) = 1 - 1 = 0 Step 3: Use the point-slope form to find the equation of the tangent. Using the point (x_1, y_1) = (1, 2) and the slope m = 0: y - 2 = 0(x - 1) y - 2 = 0 y = 2 The equation of the tangent is: y = 2 Question 49: Step 1: Determine the slope of the tangent at the given point. From question 48, the slope of the tangent m_tangent at (1, 2) is 0. Step 2: Calculate the slope of the normal. Since the tangent is a horizontal line (y=2), the normal line will be a vertical line with an undefined slope. Step 3: Find the equation of the normal. A vertical line passing through the point (1, 2) has the equation x = x_1. x = 1 The equation of the normal is: x = 1 Question 50: Step 1: Find the first derivative (dy)/(dx). Given y = sqrt(x). Rewrite as y = x^1/2. Differentiate y with respect to x: (dy)/(dx) = (1)/(2)x^(1/2)-1 = (1)/(2)x^-1/2 = (1)/(2sqrt(x)) Step 2: Evaluate the derivative at the point (4, 2) to find the slope of the tangent. Substitute x=4 into the derivative: m = (1)/(2sqrt(4)) = (1)/(2(2)) = (1)/(4) Step 3: Use the point-slope form to find the equation of the tangent. Using the point (x_1, y_1) = (4, 2) and the slope m = (1)/(4): y - 2 = (1)/(4)(x - 4) 4(y - 2) = x - 4 4y - 8 = x - 4 x - 4y + 4 = 0 The equation of the tangent is: x - 4y + 4 = 0 Last free one today — make it count tomorrow, or type /upgrade for unlimited.