Determine the vertex of y = x2 6x +19. This function is in standard form. We need to complete the square to get it to vertex form. Show using algebra tiles and with algorithm. Method 1: Method 2: Therefore the vertex is (, ). The minimum value of the function would be _when x- The min. /max. value 13 the y- portion of the vertex and where this min Imax. occurs is the x-portion of the vertex. Since the min/max. value occurs at the vertex of the function which is a turning point, the slope of the tangent (ie: the derivative) will be zero. f '(x) = O at a maximum or minimum. Therefore, this is where it occurs, thus So the max/min. occurs at x = 3 and is 10 but how do we find out whether it's a maximum or minimum without relying on previous parabola knowledge. Ie: leading coefficient is + or -. At a minimum, f '(x) 0 to the right and Vice versa for the maximum. Draw this scenario. Create a First Derivative Chart Therefore as you approach x = 3 from below it is (downward slope) and from above it is (upward slope) so f(x) has a of 10 at x = 3. In the following three graphs, the derivative equals zero at two points: y Y Y maximum maximum maximum minimum minimum minimum Eg.l: Determine the maximum or minimum values of f (x) = x3 Jc2 when x[2,l] . At a max./min. f '(x) = 0 Therefore f(x) has a local maximum at x = 0 and at x = 1 (the endpoint) and a local minimum at x = 2/3 and an absolute minimum at x = -2 (the endpoint)