 Hello and welcome to the session. In this session we shall discuss how to graph quadratic functions and show intercepts, maxima and minima. First of all we shall discuss about quadratic function. A quadratic function is a function that can be written in the form f of x is equal to a x square plus b x plus c where a cannot be equal to 0 and a b and c are constants. Now here we have to graph quadratic functions and show intercepts, maxima and minima. Now we shall learn how to determine maxima or minima. We know that the vertex is the lowest or highest point depending on direction on the graph of a quadratic function. For maxima or minima first of all we have to find the vertex of the quadratic function f of x is equal to a x square plus b x plus c. We can find the vertex by using method of completing the square and rewriting the function in the form f of x is equal to a into x minus h whole square plus k where the vertex is given by the ordered pair hk. The second method of finding the vertex is using the equation of axis of symmetry which is given by x is equal to minus b upon 2a which is the x coordinate of the vertex. Using this we can find the value of y coordinate of the vertex which is given by f of minus b upon 2a. Now if a is greater than 0 then y coordinate of the vertex is the minimum value of the function and if a is less than 0 then y coordinate of the vertex is the maximum value of the function. Now we are going to discuss about x and y intercepts. To find the y intercept we put x is equal to 0 in the given function that is y intercept is equal to f of 0. To find x intercept we solve the equation a x square plus b x plus c is equal to 0 for x. Now let us discuss an example. Let us consider an example. Here we have to graph the function given by y is equal to x square minus 5x plus 6 showing intercepts and maxima or minima. Now here we are given the function y is equal to x square minus 5x plus 6 on comparing it with standard 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 1 b is equal to minus 5 and c is equal to 6. Now we shall find its maxima or minima. Now we know that the vertex is given by the ordered pair minus b upon 2 into a f of minus b upon 2 into a. Now minus b upon 2 a will be equal to minus of minus 5 upon 2 into 1 that will be equal to 5 by 2 and f of minus b upon 2 a will be equal to 5 by 2 whole square minus 5 into 5 by 2 plus 6 that is we have put the value of x as 5 by 2 in this given function on solving this is equal to minus 1 by 4. So f of minus b upon 2 a is equal to minus 1 by 4 so the vertex is given by the ordered pair 5 upon 2 minus 1 by 4 since a is equal to 1 which is greater than 0. So y coordinate of the vertex which is equal to minus 1 by 4 is the minimum value of the function. Now we shall find the x intercept. Here we are given the quadratic function x square minus 5x plus 6 and to find the x intercept we equate this quadratic function to 0. Now this can be written as x square minus 3x minus 2x plus 6 is equal to 0. Now taking x common from first two terms we get x into x minus 3 the whole and now taking minus 2 common from last two terms and we get minus 2 into x minus 3 the whole and this is equal to 0 which implies that x minus 2 the whole into x minus 3 the whole is equal to 0 which further implies that x is equal to 2 and 3 so we say that x intercepts are given by the ordered pairs 2 0 and 3 0. Now we are going to find y intercept to find the y intercept we put x is equal to 0 in the given function that is y is equal to x square minus 5x plus 6 and we get y is equal to 0 square minus 5 into 0 plus 6 which implies that y is equal to 0 minus 5 into 0 is 0 plus 6 which further implies that y is equal to 6 so y intercept is given by the ordered pair 0 6 now let us graph this function now here is the required graph of the function y is equal to x square minus 5x plus 6 the y intercept is given by the ordered pair 0 6 and the x intercepts are given by the ordered pairs 2 0 and 3 0 and here we see that the vertex is given by the ordered pair 5 by 2 minus 1 by 4 and the y coordinate of the vertex that is minus 1 by 4 is the minimum value of the function thus in this session we have discussed how to graph quadratic functions and to show intercepts maxima and minima this completes our session hope you enjoyed this session