(a) This part of the question concerns the quadratic equation Ar' +6x+2 -0. (i) Find the discriminant of the quadratic expression da' + 6x + 2. [2] (ii) What does this discriminant tell you about the number of solutions of the equation? Explain your answer briefly. [2] (iii) What does this discriminant tell you about the graph of y - 4x- +63 +2? [1] (b) Figure 1 shows part of the graph of the quadratic function y - -3x- + 15x + 18. y (-1.0) (6,0) T. Figure 1 (i) Explain why the graph of the quadratic function y - -3x- + 15x + 18 is n-shaped. (ii) Write down the y-intercept. [1] (iii) Find the equation of the axis of symmetry of the parabola given by y - -3x + 15x + 18, explaining your method. [2] (iv) Use your answer to part (b)(iii) to find the coordinates of the vertex of the parabola given by y - -32 + 152 + 18. [2]a Here the quadratic equation is, 4# + 62+ 2 = 0 " ) Now , We are asked to find the discriminant of the quadratic expression 42 7 + 6 x + 2 Now , we know, that the discriminant of the quadratic expression an texte in, = b - yac . now , comparing our quadratic expression to ax + bute , we get , a = 4 , b = 6 and C = 2 . Hence, the discriminant of the quadratic expression, 42 + 6x + 2 is = 6 - 4 x 4x 2 = 36 - 32 = 4 . Aus ! - so , the discriminant of the quadratic expression yx + 62+2 is = 4 . ( 1' ) Now , the discriminant of the quadratic expression antby +e in = bruac. now, case-1:- If bruac = 0, then , the quadratic equation ant but c F o has two equal real roots . case - (2) . - If b-lacto, then the quadratic equation an + but c 20 has two distinct real roots. Case - 3).- If b-lac co , then the quadratic equation antbut c = o has no real roots. Now, For our Problem, the quadratic expression you"+ 6x+2 has discriminant = 4 , which is strictly greater than O. Therefore , The quadratic equation , 4 x " + 6x + 2 =0 has two distinct real roots. Now, let us find the real roots of the quadratic equation, 4 x + 6x+ 2 = 0 -) 4x + 4x + 2x + 2 = 0 Page - 1 17 (x + 2 ) ( x + 1 ) = 0 - > x5-2/4 2 = -1 So , a =-1/, and 2 5-1 Hence , the roots of the quadratic equation 4x" + 6x+ 2 = 0 are - 1 / 2 3 - 1 . "" ) Now , since , the discriminant of the quadratic expression 42 + 6x+ 2 is greater than zero. therefore , the quadratic function y = 4x " + 6x + 2 has two distinct real roots. Hence the graph of y= 42+ 6x+ 2 cuts the X-axis at two distinct real Points which are x =-1/ and x=-1. that is , the Points are , ( - 1 , 0) and ( - 12 , 0 ) . b ) Here we are given the graph of the quadratic function y = - 3 2 + 15 x+ 18 . Now, Here, First of all from the graph of y= - 32 +15x+18 cuts the x-axis at two Points ( -1 , 0 ) and ( 6 , 0 ) . NOW, 4 = - 34 + 15 x +18 = - 3 (2 -5x-6) = - 3 ( x - 6 ) ( a+ 1 ) [ as the noot of the equation is = -I and * = 67. now , when , 2 - 00 And , we know that , any quadratic function represents a Parabola . That's why the graph of the quadratic function , 9 = - 3x + 15x+ 18 in n-shaped. ( ) Now , we have to find the y-intercept of the quadratic function , y= - 3x + 15 x + 18, that is, we have to find out where the quadratic function cuts the yraxis . Now , when the function cuts the Y-axis , then the value of a is = 0. So, Putting the value x = 0, in the function, 45-3 x+15 x+18 we get , y = - 3 x 0 + 1 5 x 0 + 1 8 = 18 Hence , the quadratic function 4 = - 32 + 15 x + 18 cuts the Y- axis at (0, 18 ) Amir Therefore , the y-intercept of the quadratic function 4 = - 32 + 15 x+ 18 is =18 . (" ) Now, we have to find out the equation of the axis of symmetry . # NOW , the equation of the axis of symmetry of the Parabola cannot be yraxis , As , if we replace " x' by Ex' in the equation then the equation changes to y = - 32- 15x+ 18 . # And , Also the Parabola is not symmetric about X-axis . As , if we replace , "y' ly "- y' then the equation changes to |Page - 31- 4 =- 32 + 152+18 , that is the equation doesnot remain . same And from, the graph, we can conclude that the Parabola Cannot be symmetric about any line Parallel to x-axis. Now , Since , the Parabola cuts the X-axis at two distinct Points , that is , at the Points ( -1 , 0 ) and (6, 0 ) , therefore the Parabola will be symmetric about the line Parallel to y-axis and Passes through the mid Point of (-1, 0 ) and ( 6, 0 ) . Now, the mid Point of (-1, 0) and ( 6,0) is ( 6- 1, 6+ 0 ) that in , ( 2 . 5 , 0 ) . Hence the equation of the line Parallel to yoaxis and Passes through the mid Point of ( -1 ,0 ) and ( 6 , 0 ) is Aw ! - = 2 .5 Hence, the equation of the axis of symmetry of the Parabola is : M= 2 .5 " " ) Now, Here we have to find thes co-ordinates of the vertex of the Parabola given by ys -32+ 15x418 . Now, the vertex will be the intersection Point of the axis of symmetry and the Parabola. So, let us find the intersecting Point of 2= 2.5 and 4=-32 + 15x+18 So , 4 = - 3X ( 2 . 5 ) + 15 x 2. 5 + 18 = 36.75. Aus! " Hence, the coordinate of the vertex of the Parabola is ( 2 . 5 , 36. 75 ). [Page - 41