The graph of a function $f(x)$ is given. The function consists of a parabolic arc with vertex $V$, a line segment, and a portion of a homographic function. The domain of $f(x)$ is $\mathbb{R}$. We need to find the analytical expression for $f(x)$.

1. Problem Description

The graph of a function $f(x)$ is given. The function consists of a parabolic arc with vertex $V$, a line segment, and a portion of a homographic function. The domain of $f(x)$ is $\mathbb{R}$. We need to find the analytical expression for $f(x)$.

2. Solution Steps

Let's analyze the graph and determine the function for each part.

(a) Parabola: The vertex of the parabola is at $V = (-2, 1)$. The general form of a parabola with vertex $(h, k)$ is $y = a(x-h)^2 + k$. In our case, $h = -2$ and $k = 1$, so $y = a(x+2)^2 + 1$. The parabola is defined for $x \le -1$. When $x = -1$, we see that $y = 0$.

So, $0 = a(-1+2)^2 + 1$, which means $0 = a(1)^2 + 1$, and thus $a = -1$.

The parabola is $f(x) = -(x+2)^2 + 1 = -(x^2 + 4x + 4) + 1 = -x^2 - 4x - 3$ for $x \le -1$.

(b) Line Segment: The line segment connects the point $(-1, 0)$ to the point $(1, 1)$. The equation of a line is $y = mx + b$. We can find the slope $m = \frac{1-0}{1-(-1)} = \frac{1}{2}$. Now we can use the point-slope form $y - y_1 = m(x - x_1)$ using the point $(-1,0)$:

$y - 0 = \frac{1}{2}(x - (-1))$, so $y = \frac{1}{2}(x + 1) = \frac{1}{2}x + \frac{1}{2}$.

This line segment is defined for $-1 < x \le 1$.

(c) Homographic Function: This is a rational function of the form $f(x) = \frac{ax+b}{cx+d}$. There's a vertical asymptote at $x = 1$. This means $cx + d = 0$ when $x = 1$, so $c(1) + d = 0$, or $d = -c$. So we have $f(x) = \frac{ax+b}{cx-c}$. The function passes through $(2, 4)$.

$4 = \frac{2a+b}{2c-c} = \frac{2a+b}{c}$. So $4c = 2a+b$. As $x$ goes to infinity, the function appears to approach y=2 asymptotically. Thus $\frac{a}{c} = 2$, so $a = 2c$. Substituting $a = 2c$ into $4c = 2a+b$ gives $4c = 2(2c) + b$, which implies $4c = 4c + b$, so $b = 0$.

Thus $f(x) = \frac{2cx}{cx-c} = \frac{2x}{x-1}$ for $x > 1$.

Combining all the parts, we have:

$f(x) =

\begin{cases}

-x^2 - 4x - 3 & x \le -1 \\

\frac{1}{2}x + \frac{1}{2} & -1 < x \le 1 \\

\frac{2x}{x-1} & x > 1

\end{cases}$

3. Final Answer

$f(x) =

\begin{cases}

-x^2 - 4x - 3 & x \le -1 \\

\frac{1}{2}x + \frac{1}{2} & -1 < x \le 1 \\

\frac{2x}{x-1} & x > 1

\end{cases}$