The problem asks us to graphically determine the number of solutions to the equation $f(x) = 0$ and the solutions themselves for different functions $f(x)$ defined on specified intervals $I$. We have the following functions and intervals: a) $f(x) = x^2 - 5x + 6$, $I = [0, 5]$ b) $f(x) = -2x^2 + 12x - 18$, $I = [-4, 4]$ c) $f(x) = -x^2 + x - 5.5$, $I = [-1, 3]$ d) $f(x) = 5x^2 - 30x + 45$, $I = [-4, 6]$

1. Problem Description

The problem asks us to graphically determine the number of solutions to the equation $f(x) = 0$ and the solutions themselves for different functions $f(x)$ defined on specified intervals $I$. We have the following functions and intervals:

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

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

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

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

2. Solution Steps

a) $f(x) = x^2 - 5x + 6$. We want to solve $x^2 - 5x + 6 = 0$ on $[0, 5]$.

We can factor the quadratic as $(x - 2)(x - 3) = 0$. The solutions are $x = 2$ and $x = 3$. Both of these are in the interval $[0, 5]$.

Therefore, there are two solutions: $x=2$ and $x=3$.

b) $f(x) = -2x^2 + 12x - 18$. We want to solve $-2x^2 + 12x - 18 = 0$ on $[-4, 4]$.

Divide by $-2$ to get $x^2 - 6x + 9 = 0$. This factors as $(x - 3)^2 = 0$.

The solution is $x = 3$. Since $3$ is in the interval $[-4, 4]$, we have one solution: $x=3$.

c) $f(x) = -x^2 + x - 5.5$. We want to solve $-x^2 + x - 5.5 = 0$ on $[-1, 3]$.

Multiply by $-1$ to get $x^2 - x + 5.5 = 0$.

We use the quadratic formula to find the roots:

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

Here, $a = 1$, $b = -1$, and $c = 5.5$. So,

$x = \frac{1 \pm \sqrt{1 - 4(1)(5.5)}}{2} = \frac{1 \pm \sqrt{1 - 22}}{2} = \frac{1 \pm \sqrt{-21}}{2}$.

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

d) $f(x) = 5x^2 - 30x + 45$. We want to solve $5x^2 - 30x + 45 = 0$ on $[-4, 6]$.

Divide by $5$ to get $x^2 - 6x + 9 = 0$. This factors as $(x - 3)^2 = 0$.

The solution is $x = 3$. Since $3$ is in the interval $[-4, 6]$, we have one solution: $x=3$.

3. Final Answer

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

b) One solution: $x=3$

c) No solutions

d) One solution: $x=3$