 Hello and welcome to the session. In this session, we will compare properties of two functions each represented in a different way, that is, algebraically, graphically or by verbal description. Let us first compare properties of two functions algebraically. Now let us consider an example which says that which of the given functions has larger maximum and the functions given are f of x is equal to minus 3 into x minus 2 whole square plus 5 or g of x is equal to minus 6x square plus 24x. Now for the function f of x is equal to minus 3 into x minus 2 whole square plus 5. Now comparing this function with the vertex form of the quadratic function that is a into x minus h whole square plus k we get a is equal to minus 3, h is equal to 2 and k is equal to 5. So it's vertex which is given by the ordered pair hk will be equal to the ordered pair 25. Now as a is equal to minus 3 which is less than 0. So the graph of the function will open downwards and vertex will be the maximum point which is given by the ordered pair 25. The y coordinate of the vertex will give the maximum value of the function. So the maximum value of the function is 5. Now we take the other function that is g of x is equal to minus 6x square plus 24x. Now by method of completing the squares we get g of x is equal to now taking minus 6 common from these two terms we get minus 6 into x square minus 4x the whole which implies that g of x is equal to minus 6 into. Now adding in subtracting the square of half the coefficient of x. Now here the coefficient of x is 4. When we have this we get 2 and by squaring this coefficient that is by doing the square of 2 we get 4. So we have x square minus 4x plus 4 minus 4 the whole which further implies that g of x is equal to minus 6 into. Now x square minus 4x plus 4 can be written as x minus 2 whole square minus 4 the whole. Now here we have used the formula of a minus b whole square which is equal to a square minus 2ab plus b square. This implies that g of x is equal to minus 6 into x minus 2 whole square plus 24. Now comparing it with the vertex form of the quadratic function that is a into x minus h whole square plus k we get a is equal to minus 6 h is equal to 2 and k is equal to 24. So its vertex is given by the ordered pair hk that is the ordered pair 224. The y coordinate of the vertex will give the maximum value of the function. So the maximum value of the function is 24. So we have seen that for the function f of x is equal to minus 3 into x minus 2 whole square plus 5. The maximum value is 5 and for the function g of x which is equal to minus 6x square plus 24x the maximum value is 24. So we conclude that g of x has larger maximum value. Now let us compare the two functions graphically. Now let us consider an example. Here we are given the graph that shows two quadratic functions and we have to find the x intercepts of both the functions and we also need to compare the maxima or minima of these two functions. Now for this blue curve the x intercepts are given by the ordered pairs minus 6 0 and minus 1 0 and for red curve x intercepts are given by the ordered pairs 2 0 and 3 0. We can also see that blue curve has greater minima. Now we are going to discuss how to compare the two functions verbally. Now here let us take two cases. Here case one is the equation for the object's height h in meters and the time t seconds after launch is h of t is equal to minus 4.9 t square plus 14.7 t plus 19.6 and case two is the equation for the object's height s in feet and time t seconds after launch is s of t is equal to minus 16 t square plus 64 t plus 80. Now we have to compare the maximum height in the two cases. Now in case one the equation is given as h of t is equal to minus 4.9 t square plus 14.7 t plus 19.6. Now on comparing it with general form of the quadratic function that is f of x is equal to a x square plus b x plus c we get a is equal to minus 4.9 b is equal to 14.7 and c is equal to 19.6. Now to find maximum height we have to find the vertex and vertex is given by the ordered pair minus b upon 2a f of minus b upon 2a so minus b upon 2a will be equal to minus 14.7 upon 2 into minus 4.9 on solving this we get 3 by 2 also f of minus b upon 2a will be equal to f of 3 by 2 and here this will be equal to h of 3 by 2 and that is given by minus 4.9 into 3 by 2 whole square plus 14.7 into 3 by 2 plus 19.6 on solving this we get 30.63 so h of 3 by 2 is equal to 30.63 so vertex is given by the ordered pair 3 by 2 30.63 here since a is equal to minus of 4.9 which is less than 0 so y coordinate of the vertex which is equal to 13.63 is the maximum value of the function so we can say that for case 1 maximum height is equal to 30.63 meters now for case 2 equation given to us is s of t is equal to minus 16 t square plus 64 t plus 80 now on comparing it with general form of the quadratic function that is f of x is equal to ax square plus bx plus c we get a is equal to minus 16 b is equal to 64 and c is equal to 80 to find the maximum height we have to find the vertex and vertex is given by the ordered pair minus b upon 2a f of minus b upon 2a so minus b upon 2a will be equal to minus 64 upon 2 into minus 16 and this is equal to 2 also f of minus b upon 2a which is equal to f of 2 and here this will be equal to s of 2 is equal to minus 16 into 2 square plus 64 into 2 plus 80 on solving it we will get 144 so s of 2 is equal to 144 that is f of minus b upon 2a is equal to 144 so coordinates of the vertex is given by the ordered pair to 144 now here since a is equal to minus 16 which is less than 0 so why coordinate of the vertex which is equal to 144 is the maximum value of the function therefore for case 2 the maximum height is equal to 144 feet now to convert it in meters we multiply 144 with 0.3048 and we get it as 43.89 meters so here we have seen that for case 1 the maximum height is 30.63 meters and for case 2 the maximum height is 43.89 meters so for case 2 we have larger maximum height thus in this session we have compared property of two functions each represented in a different way that is as a brightly graphically or by verbal description this completes our session hope you enjoyed the session