Exercise $$ 2-(8 $$ points) Consider the function $$ g $$ defined over IR by: $$ g(x)=m x^{2}-(2 m-1) x+m $$ where $$ m $$ is a non-zero real parameter and designate by $$ \left(C_{m} ight) $$ its representative curve in an orthonormal system $$ (O ; \vec{l}, \vec{\jmath}) $$. 1) Show that for $$ m
eq 0,\left(C_{m} ight) $$ passes through a fixed point $$ A $$ whose coordinates are to be determined. 2) Let $$ B $$ be the symmetric of $$ A $$ with respect to the axis of symmetry of the curve $$ \left(C_{m} ight) $$. Calculate the coordinates of $$ B $$. 3) 'a) Find the slopes of the tangents at $$ A $$ and $$ B $$. b) Find the value of $$ m $$ if the tangents are perpendicular.

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Exercise $$ 2-(8 $$ points) Consider the function $$ g $$ defined over IR by: $$ g(x)=m x^{2}-(2 m-1) x+m $$ where $$ m $$ is a non-zero real parameter and designate by $$ \left(C_{m}ight) $$ its representative curve in an orthonormal system $$ (O ; \vec{l}, \vec{\jmath}) $$. 1) Show that for $$ m 
eq 0,\left(C_{m}ight) $$ passes through a fixed point $$ A $$ whose coordinates are to be determined. 2) Let $$ B $$ be the symmetric of $$ A $$ with respect to the axis of symmetry of the curve $$ \left(C_{m}ight) $$. Calculate the coordinates of $$ B $$. 3) 'a) Find the slopes of the tangents at $$ A $$ and $$ B $$. b) Find the value of $$ m $$ if the tangents are perpendicular.

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To solve the problem step by step, we will address each part of the exercise systematically.

### Part 1: Show that for $$ m \neq 0, (C_{m}) $$ passes through a fixed point $$ A $$

The function is given by:

$$
g(x) = m x^{2} - (2m - 1)x + m
$$

To find a fixed point $$ A $$ that the curve passes through for all $$ m \neq 0 $$, we can evaluate $$ g(x) $$ at a specific value of $$ x $$. A common choice is $$ x = 1 $$:

$$
g(1) = m(1)^{2} - (2m - 1)(1) + m
$$

Calculating this gives:

$$
g(1) = m - (2m - 1) + m = m - 2m + 1 + m = 1
$$

Thus, the point $$ A(1, 1) $$ is a fixed point through which the curve $$ (C_{m}) $$ passes for all $$ m \neq 0 $$.

### Part 2: Calculate the coordinates of $$ B $$

The axis of symmetry of a quadratic function $$ ax^2 + bx + c $$ is given by the formula:

$$
x = -\frac{b}{2a}
$$

For our function $$ g(x) $$:

- $$ a = m $$
- $$ b = -(2m - 1) $$

Calculating the axis of symmetry:

$$
x = -\frac{-(2m - 1)}{2m} = \frac{2m - 1}{2m}
$$

Now, we need to find the coordinates of point $$ B $$, which is the symmetric point of $$ A(1, 1) $$ with respect to the axis of symmetry. The x-coordinate of $$ B $$ can be calculated as follows:

Let $$ x_s = \frac{2m - 1}{2m} $$ be the x-coordinate of the axis of symmetry. The distance from $$ A $$ to the axis of symmetry is:

$$
d = 1 - x_s = 1 - \frac{2m - 1}{2m} = \frac{2m - (2m - 1)}{2m} = \frac{1}{2m}
$$

Thus, the x-coordinate of $$ B $$ is:

$$
x_B = x_s - d = \frac{2m - 1}{2m} - \frac{1}{2m} = \frac{2m - 1 - 1}{2m} = \frac{2m - 2}{2m} = \frac{m - 1}{m}
$$

The y-coordinate of $$ B $$ is the same as that of $$ A $$ since the reflection across the axis of symmetry does not change the y-coordinate:

$$
y_B = 1
$$

Thus, the coordinates of point $$ B $$ are:

$$
B\left(\frac{m - 1}{m}, 1\right)
$$

### Part 3: Find the slopes of the tangents at $$ A $$ and $$ B $$

To find the slopes of the tangents at points $$ A $$ and $$ B $$, we need to calculate the derivative $$ g'(x) $$:

$$
g'(x) = \frac{d}{dx}(m x^{2} - (2m - 1)x + m) = 2mx - (2m - 1)
$$

Now, we evaluate the derivative at point $$ A(1, 1) $$:

$$
g'(1) = 2m(1) - (2m - 1) = 2m - 2m + 1 = 1
$$

So, the slope of the tangent at $$ A $$ is $$ 1 $$.

Next, we evaluate the derivative at point $$ B\left(\frac{m - 1}{m}, 1\right) $$:

$$
g'\left(\frac{m - 1}{m}\right) = 2m\left(\frac{m - 1}{m}\right) - (2m - 1) = \frac{2m(m - 1)}{m} - (2m - 1) = 2(m - 1) - (2m - 1) = 2m - 2 - 2m + 1 = -1
$$

So, the slope of the tangent at $$ B $$ is $$ -1 $$.

### Part 3b: Find the value of $$ m $$ if the tangents are perpendicular

Two lines are perpendicular if the product of their slopes is $$ -1 $$. The slopes we found are $$ 1 $$ (at $$ A $$) and $$ -1 $$ (at $$ B $$).

Calculating the product:

$$
1 \cdot (-1) = -1
$$

Since the product is already $$ -1 $$, the tangents at points $$ A $$ and $$ B $$ are perpendicular for any non-zero value of $$ m $$.

### Summary of Results

1. The fixed point $$ A $$ is $$ (1, 1) $$.
2. The coordinates of $$ B $$ are $$ \left(\frac{m - 1}{m}, 1\right) $$.
3. The slopes of the tangents at $$ A $$ and $$ B $$ are $$ 1 $$ and $$ -1 $$, respectively, and they are perpendicular for any non-zero $$ m $$.
For any non-zero $$ m $$, the curve $$ (C_{m}) $$ passes through the fixed point $$ A(1, 1) $$. The symmetric point $$ B $$ has coordinates $$ \left(\frac{m - 1}{m}, 1\right) $$. The slopes of the tangents at $$ A $$ and $$ B $$ are $$ 1 $$ and $$ -1 $$, respectively, making them perpendicular for all non-zero $$ m $$.

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