The diagram on the right shows the graph of a quadratic function, $f$ - Leaving Cert Mathematics - Question 5 - 2018

To determine the quadratic function, we can plug known points into the standard form $f(x) = ax^2 + bx + c$.

Using C (3, 0):

o = 9a + 3b + c Since c = 6, substituting gives:

o = 9a + 3b + 6

Using B (-1, 5):

5 = a(-1)^2 + b(-1) + 6 This simplifies to:

5 = a - b + 6 Thus:

-1 = a - b

Now use A (-2, 6):

6 = a(-2)^2 + b(-2) + c = 4a - 2b + 6 This leads to:

0 = 4a - 2b

Solving the two equations-1 = a - b and 0 = 4a - 2b: We can write 2b = 4a Then substituting gives: a = 1/2, b = 2 Therefore, $f(x) = -x^2 + x + 6$.

First, we differentiate the function: $f'(x) = -2x + 1$ Setting the derivative to zero to find critical points: $-2x + 1 = 0$ Solving for x gives: x = rac{1}{2}

Now, to find the y-coordinate, substitute x = rac{1}{2} back into the function: f(0.5) = -rac{1}{4} + rac{1}{2} + 6 = 6 - rac{1}{4} = rac{24 - 1}{4} = rac{23}{4} = 6.25 Thus, the maximum point is (0.5, 6.25).