The problem consists of two exercises. Exercise 2 deals with a linear function $f$ such that $f(7) = 14$. We need to find the expression of $f(x)$, find the number $a$ such that $f(a) = 21$, and draw the graph of $f$. Exercise 3 deals with an affine function $g$ such that $g(0) = 3$ and $g(2) = 7$. We need to find the expression of $g(x)$, calculate $g(1)$ and $g(3)$, find the number $a$ such that $g(a) = 21$, and draw the graph of $g$. Note that the image asks for the number $a$ such that $g(a) = 21$ by the function $f$ - but the problem is likely asking for the value by the function $g$.

1. Problem Description

The problem consists of two exercises. Exercise 2 deals with a linear function $f$ such that $f(7) = 14$. We need to find the expression of $f(x)$, find the number $a$ such that $f(a) = 21$, and draw the graph of $f$. Exercise 3 deals with an affine function $g$ such that $g(0) = 3$ and $g(2) = 7$. We need to find the expression of $g(x)$, calculate $g(1)$ and $g(3)$, find the number $a$ such that $g(a) = 21$, and draw the graph of $g$. Note that the image asks for the number $a$ such that $g(a) = 21$ by the function $f$ - but the problem is likely asking for the value by the function $g$.

2. Solution Steps

Exercise 2:

(1) Since $f$ is a linear function, it has the form $f(x) = kx$ for some constant $k$. We are given that $f(7) = 14$, so $7k = 14$.

$k = \frac{14}{7} = 2$

Therefore, $f(x) = 2x$.

(2) We want to find $a$ such that $f(a) = 21$. This means $2a = 21$.

$a = \frac{21}{2} = 10.5$

(3) To draw the line $f(x) = 2x$, we need two points. We know that $f(0) = 0$ and $f(7) = 14$. So the line passes through $(0, 0)$ and $(7, 14)$.

Exercise 3:

(1) Since $g$ is an affine function, it has the form $g(x) = ax + b$ for some constants $a$ and $b$.

We are given that $g(0) = 3$ and $g(2) = 7$.

$g(0) = a(0) + b = 3$, so $b = 3$.

$g(2) = a(2) + b = 7$, so $2a + 3 = 7$. Therefore, $2a = 4$ and $a = 2$.

Thus, $g(x) = 2x + 3$.

(2) $g(1) = 2(1) + 3 = 2 + 3 = 5$

$g(3) = 2(3) + 3 = 6 + 3 = 9$

(3) We want to find $a$ such that $g(a) = 21$. This means $2a + 3 = 21$.

$2a = 21 - 3 = 18$

$a = \frac{18}{2} = 9$

(4) To draw the line $g(x) = 2x + 3$, we need two points. We know that $g(0) = 3$ and $g(2) = 7$. So the line passes through $(0, 3)$ and $(2, 7)$.

3. Final Answer

Exercise 2:

(1) $f(x) = 2x$

(2) $a = 10.5$

(3) The graph of $f$ passes through $(0, 0)$ and $(7, 14)$.

Exercise 3:

(1) $g(x) = 2x + 3$

(2) $g(1) = 5$ and $g(3) = 9$

(3) $a = 9$

(4) The graph of $g$ passes through $(0, 3)$ and $(2, 7)$.