The problem asks us to determine the number of solutions and the solutions of the equation $f(x) = 0$ graphically, for each given function $f(x)$ within a specified interval $I$. 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 determine the number of solutions and the solutions of the equation $f(x) = 0$ graphically, for each given function $f(x)$ within a specified interval $I$.

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 need to find the roots of $f(x)=0$ in the interval $[0, 5]$.

$x^2 - 5x + 6 = 0$

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

$x = 2$ or $x = 3$

Both roots are in the interval $[0, 5]$.

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

b) $f(x) = -2x^2 + 12x - 18$

We need to find the roots of $f(x)=0$ in the interval $[-4, 4]$.

$-2x^2 + 12x - 18 = 0$

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

$(x - 3)^2 = 0$

$x = 3$

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

Therefore, there is 1 solution: $x = 3$.

c) $f(x) = -x^2 + x - 5.5$

We need to find the roots of $f(x)=0$ in the interval $[-1, 3]$.

$-x^2 + x - 5.5 = 0$

$x^2 - x + 5.5 = 0$

We can use the quadratic formula to find the roots:

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

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

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

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

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

Therefore, there are 0 solutions in the interval $[-1, 3]$.

d) $f(x) = 5x^2 - 30x + 45$

We need to find the roots of $f(x)=0$ in the interval $[-4, 6]$.

$5x^2 - 30x + 45 = 0$

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

$(x - 3)^2 = 0$

$x = 3$

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

Therefore, there is 1 solution: $x = 3$.

3. Final Answer

a) Number of solutions: 2, Solutions: $x=2$, $x=3$

b) Number of solutions: 1, Solution: $x=3$

c) Number of solutions: 0

d) Number of solutions: 1, Solution: $x=3$