A curve has the equation y = (1/3)x^3 + 4x^2 + 12x +3. Find the coordinates of each turning point and determine their nature.

  The turning point of a curve can be visualised in your head (or on paper/screen using diagram) as the point in which the curve charges direction – and at this point the gradient of the curve is equal to 0 as it is completely horizontal – so there is no change in y by change in x. By knowing this, we can say that dy/dx = 0. To start this question off, we can find dy/dx before equating it to 0. By doing this and differentiating the above equation we can say that dy/dx = x^2 +8x + 12 = 0. Now this looks like a simple quadratic. As we have seen in GCSE maths, a quadratic equated to 0 can be solved, and by solving this we can get the x-coordinates of the turning points. This then becomes (x+6)(x+2) = 0, and we can say that x = -6 or x = -2.
   To find the y coordinate we can these into the original equation of the curve. By doing this, we get y= 3 (for x =-6) and y =-7.67 (for x = -2). Therefore our coordinates are (-6, 3) and (-2, -7.67). In order to determine the nature of these points, we must find the second differential and sub in the values of x. d2y/dx2 = 2x+8. Subbing in for the first coordinate gives us d2y/dx2 = -4 which is less than zero. This means that (-6, 3) is a maximum turning point. Subbing in for the second coordinate gives us d2y/dx2 = 4 therefore (-2, -7.67) is a minimum turning point. Remember, whenever the second differential is greater than 0, the turning point is greater than 0 (d2y/dx2 > 0) then it is a minimum, d2y/dx2 < 0 is a maximum. For d2y/dx2 = 0, the point could be max, min or point of inflexion. As a way of good habits, always finish your answer off by stating clearly the coordinate and it’s nature at the end of your writing, so it is easy for the examiner to see your answer.