 Hello and welcome to the session. In this session we will discuss a question which says that complete square to rewrite the given quadratic equation in what x form then determine the extreme point and x is of symmetry and sketch the graph of the equation y is equal to minus 3x square plus 30x minus 71. Now let us start with the solution of the given question. Now we have given the equation y is equal to minus 3x square plus 30x minus 71. Now we have to complete its square. Now for completing its square first of all we will make coefficient of x square 1. So we will take minus 3 common. Now taking minus 3 common we have y is equal to minus 3 into x square minus 10x plus 71 upon 3 the whole. Now here you can see that coefficient of x is minus 10 and its half is 1 upon 2 into minus 10 which is equal to minus 5 and square of minus 5 is equal to minus 5 whole square that is equal to 25. So we add and subtract square of half the coefficient of x that is 25 to make perfect square. Now we have y is equal to minus 3 into x square minus 10x plus 71 upon 3 plus 25 minus 25 the whole. This implies y is equal to minus 3 into. Now we will combine these three terms. Now we will write x square minus 10x plus 25 the whole and we will combine these two terms. So it will be plus 71 upon 3 minus 25 the whole and this complete whole. This implies y is equal to minus 3 into x square minus 10x can be written as 2 into x into 5 plus 25 can be written as 5 square the whole plus 71 upon 3 minus 25 the whole and this complete whole. Now we know that a square minus 2 into a into b plus b square or simply we can say a square minus 2 a b plus b square is equal to a minus 3 whole square. Now we will apply this formula here. Now this implies y is equal to minus 3 into now using this formula here we have x minus 5 whole square plus here we will take the lcm so the denominator we have 3 and in denominator we have 71 minus 75 the whole and this complete whole. Now this implies y is equal to minus 3 into x minus 5 whole square minus 71 minus 75 is minus 4 and plus minus 4 upon 3 will be minus 4 upon 3 the whole. Now this implies y is equal to now we will open this square bracket and we have minus 3 into x minus 5 whole square minus 3 into minus 4 upon 3 will be plus 4. So we have made its perfect square. Now we know that equation of quadratic function in vertex form is given by y is equal to a into x minus 5 whole square plus 2 where coordinates of vertex are given by h2 and x is of symmetry is given by the equation x is equal to h. Now let us compare this equation with the equation in vertex form. Now comparing it with vertex form we have a is equal to minus 3, h is equal to 5 and k is equal to 4 coordinates of vertex are given by 4 and axis of symmetry is given by the equation x is equal to h that is x is equal to 5. Now we also know if a is less than 0 then graph will open downward and thus vertex is maximum point and if a is greater than 0 then graph will open upward and thus vertex is minimum point. Now here we can see we have a is equal to minus 3 which is less than 0. So graph will be downward facing parabola and vertex is maximum point. Now let us plot the vertex that is the point with coordinates 5, 4 on the graph. So this is the point with coordinates 5, 4. Now let us show the axis of symmetry that is the line with equation x is equal to 5. So this is the vertical line whose equation is x is equal to 5. Now let this equation be equation number 1. Now let us find one additional point for this. Let us take x is equal to 6. So we put x is equal to 6 in equation 1 and we have y is equal to minus 3 into 6 minus 5 whole square plus 4 which implies y is equal to minus 3 into, now 6 minus 5 is 1 and 1 square is 1. So we have y is equal to minus 3 into 1 plus 4 which implies y is equal to minus 3 plus 4. So we have y is equal to 1. So for x is equal to 6 we have y is equal to 1. So the point with coordinates 6, 1 lies on the curve. Now let us plot this additional point on the graph. So this is the point with coordinates 6, 1. Now let us join the point with coordinates 5, 4 and the point with coordinates 6, 1 using free mode curve moving downwards. So using axis of symmetry we have drawn similar curve on other side of the axis of symmetry and we have extended this curve downwards from both ends. So this is the required curve and this is the solution of the given question. That is all for this session. Hope you all have enjoyed this session.