The problem asks us to graphically determine the number and solutions of the equation $f(x) = 0$ for the given functions $f(x)$ within the specified intervals $I$. This means we need to find the x-values where the graph of each function intersects the x-axis ($y=0$) within the given interval. Since we don't have the graphs, we will solve each equation algebraically, and then check if the solutions lie within the given intervals.

1. Problem Description

The problem asks us to graphically determine the number and solutions of the equation $f(x) = 0$ for the given functions $f(x)$ within the specified intervals $I$. This means we need to find the x-values where the graph of each function intersects the x-axis ($y=0$) within the given interval. Since we don't have the graphs, we will solve each equation algebraically, and then check if the solutions lie within the given intervals.

2. Solution Steps

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

To find the solutions of $f(x) = 0$, we need to solve the quadratic equation $x^2 - 5x + 6 = 0$.

We can factor the quadratic as $(x - 2)(x - 3) = 0$.

This gives us two solutions: $x = 2$ and $x = 3$.

Both solutions, $2$ and $3$, are within the interval $[0, 5]$.

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

To find the solutions of $f(x) = 0$, we need to solve the quadratic equation $-2x^2 + 12x - 18 = 0$.

We can divide by $-2$ to simplify the equation: $x^2 - 6x + 9 = 0$.

We can factor the quadratic as $(x - 3)(x - 3) = 0$.

This gives us one solution: $x = 3$.

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

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

To find the solutions of $f(x) = 0$, we need to solve the quadratic equation $-x^2 + x - 5.5 = 0$.

Multiply by -1: $x^2 - x + 5.5 = 0$

We can use the quadratic formula to find the solutions:

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

where $a = 1$, $b = -1$, and $c = 5.5$.

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

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

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

Since the discriminant is negative, there are no real solutions. Thus, there are no real roots.

Therefore, there are no solutions within the interval $[-1, 3]$.

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

To find the solutions of $f(x) = 0$, we need to solve the quadratic equation $5x^2 - 30x + 45 = 0$.

We can divide by $5$ to simplify the equation: $x^2 - 6x + 9 = 0$.

We can factor the quadratic as $(x - 3)(x - 3) = 0$.

This gives us one solution: $x = 3$.

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

3. Final Answer

a) Two solutions: $x = 2, 3$

b) One solution: $x = 3$

c) No real solutions.

d) One solution: $x = 3$