The problem asks us to determine the number of solutions and the solutions themselves of the equation $f(x) = 0$ for each given function $f(x)$ over the specified interval $I$. The problem states that this should be done graphically, but since we don't have the graph, we will solve the problem algebraically.

1. Problem Description

The problem asks us to determine the number of solutions and the solutions themselves of the equation $f(x) = 0$ for each given function $f(x)$ over the specified interval $I$. The problem states that this should be done graphically, but since we don't have the graph, we will solve the problem algebraically.

2. Solution Steps

a) $f(x) = x^2 - 5x + 6$ ; $I = [0; 5]$

We need to solve $x^2 - 5x + 6 = 0$. This is a quadratic equation. We can factor it:

$(x - 2)(x - 3) = 0$

So, $x = 2$ or $x = 3$. Both solutions are in the interval $[0, 5]$.

b) $f(x) = -2x^2 + 12x - 18$; $I = [-4; 4]$

We need to solve $-2x^2 + 12x - 18 = 0$. Divide by $-2$:

$x^2 - 6x + 9 = 0$

$(x - 3)^2 = 0$

So, $x = 3$. The solution is in the interval $[-4, 4]$.

c) $f(x) = -x^2 + x - 5.5$; $I = [-1; 3]$

We need to solve $-x^2 + x - 5.5 = 0$. Multiply by $-1$:

$x^2 - x + 5.5 = 0$

We use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(5.5)}}{2(1)}$

$x = \frac{1 \pm \sqrt{1 - 22}}{2}$

$x = \frac{1 \pm \sqrt{-21}}{2}$

Since the discriminant is negative, there are no real solutions. Therefore, there are no solutions in the interval $[-1, 3]$.

d) $f(x) = 5x^2 - 30x + 45$; $I = [-4; 6]$

We need to solve $5x^2 - 30x + 45 = 0$. Divide by $5$:

$x^2 - 6x + 9 = 0$

$(x - 3)^2 = 0$

So, $x = 3$. The solution is in the interval $[-4, 6]$.

3. Final Answer

a) 2 solutions: $x = 2$ and $x = 3$

b) 1 solution: $x = 3$

c) 0 solutions

d) 1 solution: $x = 3$