Given the function $f(x) = \frac{ax^2 + bx + c}{x - 2}$, we need to determine the values of $a$, $b$, and $c$ such that the curve (C) representing $f(x)$ satisfies the following conditions: * (C) passes through the point A(0, 5). * The tangent to (C) at point A is parallel to the x-axis. * The tangent to (C) at the point B with x-coordinate 1 has a slope of -3.

1. Problem Description

Given the function $f(x) = \frac{ax^2 + bx + c}{x - 2}$, we need to determine the values of $a$, $b$, and $c$ such that the curve (C) representing $f(x)$ satisfies the following conditions:

* (C) passes through the point A(0, 5).

* The tangent to (C) at point A is parallel to the x-axis.

* The tangent to (C) at the point B with x-coordinate 1 has a slope of -

3.

2. Solution Steps

* Condition 1: (C) passes through A(0, 5)

This means $f(0) = 5$. Substituting $x = 0$ into the function:

$f(0) = \frac{a(0)^2 + b(0) + c}{0 - 2} = \frac{c}{-2}$

So, $\frac{c}{-2} = 5$, which implies $c = -10$.

* Condition 2: The tangent at A(0, 5) is parallel to the x-axis.

This means $f'(0) = 0$. Let's find the derivative $f'(x)$:

$f'(x) = \frac{(2ax + b)(x - 2) - (ax^2 + bx + c)(1)}{(x - 2)^2}$

$f'(x) = \frac{2ax^2 - 4ax + bx - 2b - ax^2 - bx - c}{(x - 2)^2}$

$f'(x) = \frac{ax^2 - 4ax - 2b - c}{(x - 2)^2}$

Now, we set $f'(0) = 0$:

$f'(0) = \frac{a(0)^2 - 4a(0) - 2b - c}{(0 - 2)^2} = \frac{-2b - c}{4} = 0$

This implies $-2b - c = 0$. Since $c = -10$, we have $-2b - (-10) = 0$, which gives $-2b + 10 = 0$, so $b = 5$.

* Condition 3: The tangent at B(1, f(1)) has a slope of -

3. This means $f'(1) = -3$. We have $f'(x) = \frac{ax^2 - 4ax - 2b - c}{(x - 2)^2}$. Substituting $x = 1$, $b = 5$, and $c = -10$:

$f'(1) = \frac{a(1)^2 - 4a(1) - 2(5) - (-10)}{(1 - 2)^2} = \frac{a - 4a - 10 + 10}{1} = -3a$

So, $-3a = -3$, which implies $a = 1$.

Therefore, $a = 1$, $b = 5$, and $c = -10$.

3. Final Answer

$a = 1$, $b = 5$, $c = -10$