For the curve y = -2x3 + 3x2, find the x-intercepts, y-intercept, and stationary points.

You're on a roll — here's the solution to question 23. The equation of the curve is y = -2x^3 + 3x^2. *a) Find: (i) the x-intercept of the curve. Step 1: Set y=0 to find the x-intercepts. 0 = -2x^3 + 3x^2 Step 2: Factor out x^2. 0 = x^2(-2x + 3) Step 3: Solve for x. x^2 = 0 x = 0 -2x + 3 = 0 2x = 3 x = (3)/(2) The x-intercepts are (0, 0) and ((3)/(2), 0). (ii) the y-intercept of the curve. Step 1: Set x=0 to find the y-intercept. y = -2(0)^3 + 3(0)^2 Step 2: Calculate y. y = 0 The y-intercept is (0, 0). *b) (i) Determine the stationary points of the curve. Step 1: Find the first derivative (dy)/(dx). y = -2x^3 + 3x^2 (dy)/(dx) = -6x^2 + 6x Step 2: Set (dy)/(dx) = 0 to find the x-coordinates of the stationary points. -6x^2 + 6x = 0 Step 3: Factor out -6x. -6x(x - 1) = 0 Step 4: Solve for x. -6x = 0 x = 0 x - 1 = 0 x = 1 Step 5: Substitute the x-values back into the original equation y = -2x^3 + 3x^2 to find the corresponding y-values. For x=0: y = -2(0)^3 + 3(0)^2 = 0 Point: (0, 0) For x=1: y = -2(1)^3 + 3(1)^2 = -2 + 3 = 1 Point: (1, 1) The stationary points are (0, 0) and (1, 1). (ii) For each point in (b)(i) above, determine whether it is a maximum or a minimum. Step 1: Find the second derivative (d^2y)/(dx^2). (dy)/(dx) = -6x^2 + 6x (d^2y)/(dx^2) = -12x + 6 Step 2: Evaluate (d^2y)/(dx^2) at each stationary point. For (0, 0) (where x=0): (d^2y)/(dx^2) = -12(0) + 6 = 6 Since (d^2y)/(dx^2) > 0, the point (0, 0) is a local minimum. For (1, 1) (where x=1): (d^2y)/(dx^2) = -12(1) + 6 = -12 + 6 = -6 Since (d^2y)/(dx^2) < 0, the point (1, 1) is a local maximum. *c) Sketch the curve. The curve has the following characteristics: • x-intercepts: (0, 0) and ((3)/(2), 0) • y-intercept: (0, 0) • Local minimum: (0, 0) • Local maximum: (1, 1) The sketch should show a curve that: 1. Comes from positive y-values (as x -, y ). 2. Decreases to a local minimum at (0, 0). 3. Increases to a local maximum at (1, 1). 4. Decreases again, passing through the x-axis at ((3)/(2), 0) and continuing towards negative y-values (as x , y -). Here is a description of the sketch: Draw a coordinate plane. Plot the points (0,0), (1,1), and ((3)/(2),0). The curve starts from the top-left, goes down to touch the origin (0,0) as a minimum point. From the origin, it turns and goes up to the point (1,1) as a maximum point. From (1,1), it turns and goes down, crossing the x-axis at ((3)/(2),0) and continuing downwards. Got more? Send 'em!