76 a x2 C1 2 C1 for clear explanation of what equation to look for. H If the coefficient of is positive, the P1 P1 for choosing a suitable equation with these turning point is between the two roots, characteristics. so choose an equation with two x x x positive roots, say = 1 and = 3. The quadratic that has these roots is y = (x 1)(x 3), which is – – y x2 = 4x + 3. – y Complete the square to get = (x – 2)2 22 + 3. – y Hence the turning point of = (x 2)2 – x-value. 1 will have a positive – b C1 C1 for a suitable equation with complete justification C1 C1 for clear explanation of what equation to look for. If the equation has no roots, then the x-axis, turning point will be above the P1 P1 for choosing a suitable equation with these y. hence a positive value of characteristics. From the general form of the quadratic y ax2 bx c, equation, = + + the value b2 a of is less than 4ac so, keeping as c b 1, we could choose as 6 and as 2, y x2 giving = + 2x + 6 y Complete the square to get = (x + 1)2 1 + 6. – C1 C1 for a suitable equation with complete justification y c Hence the turning point of = (x + 1)2 P1 P1 for complete clear explanation + 5 will have a positive y value. y-intercept The will be positive if is y x positive when = 0. for example: 7 y x y = (x + 2)2 + 3, when = 0 = 7, y positive so = (x + 2)2 + 3 has a that is positive.
76 a b c If the coefficient of x2 is positive, the turning point is between the two roots, so choose an equation with two positive x roots, say x = 1 and x = 3. The quadratic that has these roots is y = (x – 1)(x – 3), which is y = x2 – 4x + 3. Complete the square to get y = (x – 2)2 – 22 + 3. Hence the turning point of y = (x – 2)2 – 1 will have a positive x-value. If the equation has no roots, then the turning point will be above the x-axis, hence a positive value of y. From the general form of the quadratic equation, y = ax2 + bx + c, the value of b2 is less than 4ac so, keeping a as 1, we could choose c as 6 and b as 2, giving y = x2 + 2x + 6 Complete the square to get y = (x + 1)2 – 1 + 6. Hence the turning point of y = (x + 1)2 + 5 will have a positive y value. The y-intercept will be positive if y is positive when x = 0. for example: y = (x + 2)2 + 3, when x = 0 y = 7, positive so y = (x + 2)2 + 3 has a that is positive. C1 P1 C1 C1 P1 C1 P1 2 C1 for clear explanation of what equation to look for. P1 for choosing a suitable equation with these characteristics. C1 for a suitable equation with complete justification C1 for clear explanation of what equation to look for. P1 for choosing a suitable equation with these characteristics. C1 for a suitable equation with complete justification P1 for complete clear explanation H 7