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

Starting with the standard form of a quadratic function:

$f(x) = ax^2 + bx + c$

From the graph, we can substitute points A, B, and C to find the constants:

1. For point A (0, 6): $c = 6$

2. For point B (-2, 0): $0 = (-2)^2 a + (-2)b + 6$ Simplifying gives: $0 = 4a - 2b + 6$ Rearranging we have:

   $$$$

ightarrow 4a - 2b = -6$$ (Equation 1)

3. For point C (3, 0): $0 = (3)^2 a + (3)b + 6$ Simplifying gives: $0 = 9a + 3b + 6$ Rearranging we have: $$$$

ightarrow 9a + 3b = -6$$ (Equation 2)

Setting up the equations:

1. From Equation 1: $4a - 2b = -6$

2. From Equation 2: $9a + 3b = -6$

Solving the two equations simultaneously, we find: $a = -1, b = 1$

Thus, we have the function: $f(x) = -x^2 + x + 6$

To find the maximum, we first take the derivative of the function:

$f(x) = -x^2 + x + 6$

The derivative is:

$f'(x) = -2x + 1$

Setting the derivative to zero for critical points:

$-2x + 1 = 0$

Solving for x gives:

ightarrow x = rac{1}{2}$$ Now, substitute $x = 0.5$ back into the original function to find the maximum value: $$f(0.5) = - (0.5)^2 + (0.5) + 6$$ Calculating: $$f(0.5) = -0.25 + 0.5 + 6 = 6.25$$ Therefore, the maximum point of $f(x)$ is (0.5, 6.25).