Unit 3- Activity 2MCV4UMinimum and Maximum on an Interval (Extreme Values)Determine the vertex of y=x2-6x19. 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 is the y-portion of the vertex and where this min./max. occurs is the x-portion of the vertex.Calculus: Determine the max./min. value of f(x)=x2-6x19 and state where it occurs.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)=0 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 left of f'(x) and>0 to the right and vice versa for the maximum. Draw this scenario.Create a First Derivative Chart\table[[Interval,f'(x)=2x-6