We are given a function $f$ and its graph $(D)$. We need to determine the nature of the function $f$, find the image of $3$ by the function $f$, and determine the expression of $f(x)$. Then we are given an affine function $g$ and its graph $(\Delta)$. We know that $g(3)-g(-1) = 6$ and $A(-1; 5) \in (\Delta)$. We need to show that $g(x) = \frac{3}{2}x + \frac{13}{2}$. We then need to calculate $g(5)$ and $g(2025) - g(2020)$. After that, we have to calculate the number whose image is 11 by the function $g$. Finally, we calculate $m$ knowing that $B(m+2; 8) \in (\Delta)$. In part 3, we have to show that the lines $(D)$ and $(\Delta)$ are perpendicular in $M(-3; 2)$.

1. Problem Description

We are given a function $f$ and its graph $(D)$. We need to determine the nature of the function $f$, find the image of $3$ by the function $f$, and determine the expression of $f(x)$. Then we are given an affine function $g$ and its graph $(\Delta)$. We know that $g(3)-g(-1) = 6$ and $A(-1; 5) \in (\Delta)$. We need to show that $g(x) = \frac{3}{2}x + \frac{13}{2}$. We then need to calculate $g(5)$ and $g(2025) - g(2020)$. After that, we have to calculate the number whose image is 11 by the function $g$. Finally, we calculate $m$ knowing that $B(m+2; 8) \in (\Delta)$. In part 3, we have to show that the lines $(D)$ and $(\Delta)$ are perpendicular in $M(-3; 2)$.

2. Solution Steps

Part 1:

a) The graph of $f$ is a straight line, so $f$ is an affine function.

b) From the graph, we can see that when $x=3$, $f(x) = -3/2*3 + 0.5 = -3$ approximately. So $f(3) = -3$. We are reading the value of $f(3)$ from the given graph. Looking at the graph we can see that the line passes through the point $(0, 0.5)$ and the gradient is $-1/2$. When $x=3, y$ is near to $-1$. The point on the graph at $x=3$ is approximately at $y=-1$.

c) From the graph, we can see that the line passes through $(0, 0.5)$ and $(1, 0)$. The slope of the line is $(0-0.5)/(1-0) = -0.5 = -1/2$. Thus $f(x) = -\frac{1}{2}x + b$. Since the line passes through $(0, 0.5)$, we have $f(0) = -\frac{1}{2}(0) + b = 0.5$, so $b = 0.5$. Hence $f(x) = -\frac{1}{2}x + \frac{1}{2}$.

Part 2:

a) $g(x)$ is an affine function, so $g(x) = ax + b$.

$g(3) - g(-1) = (3a + b) - (-a + b) = 4a = 6$, so $a = \frac{6}{4} = \frac{3}{2}$.

Since $A(-1; 5) \in (\Delta)$, $g(-1) = 5$. Thus $g(-1) = a(-1) + b = -\frac{3}{2} + b = 5$, so $b = 5 + \frac{3}{2} = \frac{10}{2} + \frac{3}{2} = \frac{13}{2}$.

Therefore $g(x) = \frac{3}{2}x + \frac{13}{2}$.

b) $g(5) = \frac{3}{2}(5) + \frac{13}{2} = \frac{15}{2} + \frac{13}{2} = \frac{28}{2} = 14$.

$g(2025) - g(2020) = \frac{3}{2}(2025) + \frac{13}{2} - (\frac{3}{2}(2020) + \frac{13}{2}) = \frac{3}{2}(2025 - 2020) = \frac{3}{2}(5) = \frac{15}{2}$.

c) We want to find $x$ such that $g(x) = 11$.

$\frac{3}{2}x + \frac{13}{2} = 11$

$\frac{3}{2}x = 11 - \frac{13}{2} = \frac{22}{2} - \frac{13}{2} = \frac{9}{2}$

$x = \frac{9}{2} \cdot \frac{2}{3} = \frac{9}{3} = 3$.

d) We know that $B(m+2; 8) \in (\Delta)$, so $g(m+2) = 8$.

$g(m+2) = \frac{3}{2}(m+2) + \frac{13}{2} = 8$

$\frac{3}{2}(m+2) = 8 - \frac{13}{2} = \frac{16}{2} - \frac{13}{2} = \frac{3}{2}$

$m+2 = \frac{3}{2} \cdot \frac{2}{3} = 1$

$m = 1 - 2 = -1$.

Part 3:

The slope of line $(D)$ is $-\frac{1}{2}$.

The slope of line $(\Delta)$ is $\frac{3}{2}$.

The product of the slopes is $-\frac{1}{2} \cdot \frac{3}{2} = -\frac{3}{4} \neq -1$, so the lines are not perpendicular.

Now, we have to check if the point $M(-3; 2)$ lies on both lines.

For line $(D)$, $f(-3) = -\frac{1}{2}(-3) + \frac{1}{2} = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$, so $M(-3; 2)$ is on line $(D)$.

For line $(\Delta)$, $g(-3) = \frac{3}{2}(-3) + \frac{13}{2} = -\frac{9}{2} + \frac{13}{2} = \frac{4}{2} = 2$, so $M(-3; 2)$ is on line $(\Delta)$.

However, the product of the slopes is not equal to -1, so the lines aren't perpendicular.

3. Final Answer

Part 1:

a) $f$ is an affine function.

b) $f(3) = -1$. Based on the graph given.

c) $f(x) = -\frac{1}{2}x + \frac{1}{2}$.

Part 2:

a) $g(x) = \frac{3}{2}x + \frac{13}{2}$.

b) $g(5) = 14$ and $g(2025) - g(2020) = \frac{15}{2}$.

c) $3$.

d) $m = -1$.

Part 3:

The lines are not perpendicular.