Sheri has been asked to find the quadratic equation of the g

Sheri has been asked to find the quadratic equation of the graph below. She can tell the solutions are x = 3 and x = 6. She also knows that the parabola opens up so the value of “a” would be positive. Because the axis of symmetry isn’t through a whole number she can’t figure out where to start. Help Sheri figure out the equation of the quadratic. Do this using algebra and give a reason with each step you need to take

Solution

Let the equation of the parabola be y = ax² + bx + c where a is positive so that the parabolaopens upwards. Since x = 3 is a zero, we have 0 = a(3)² + b(3) + c or, 9a + 3b + c = 0..(1). Further,  since x = 6 is also a zero, we have 0 =a(6)² +b(6) + c or, 36a + 6b + c = 0 ...(2) On subtracting the 1st equation from the 2nd equation, we get 36a + 6b +c - 9a - 3b - c = 0 or, 27a + 3b = 0 or, on dviding both the sides by 3, we have 9a + b = 0 or, b = -9a. On substituting b = -9a in the equation of the parabola, we get y = ax² -9ax + c = 0 = a[ x² - 2*9/2x +(9/2)² ] + c -a(9/2)² = a(x - 9/2)² + ( c - 81/4 a). This is the equation of a parabola which opens upwards ( as a is positive) and has its vertex at ( 9/2, c - 81/4a) where a, c are arbitrary constants, with the constraint that a is positive. The axis of symmetry is the line x = 9/2.