The problem asks to find the analytical expression of the function $f(x)$ whose graph is shown. The graph is composed of three parts: a parabola, a segment, and a hyperbola.

1. Problem Description

The problem asks to find the analytical expression of the function $f(x)$ whose graph is shown. The graph is composed of three parts: a parabola, a segment, and a hyperbola.

2. Solution Steps

First, let's analyze the 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 this case, $h = -2$ and $k = 1$, so the equation is $y = a(x+2)^2 + 1$.

The parabola is defined for $x \le 0$.

At $x = -1$, the parabola has a value of $0$.

$0 = a(-1+2)^2 + 1$

$0 = a(1)^2 + 1$

$a = -1$

So the parabola equation is $y = -(x+2)^2 + 1 = -(x^2 + 4x + 4) + 1 = -x^2 - 4x - 3$ for $x \le -1$.

Next, let's analyze the segment.

The segment connects the points $(-1, 0)$ and $(1, 1)$.

The slope of the segment is $m = \frac{1-0}{1-(-1)} = \frac{1}{2}$.

The equation of the line is $y - 0 = \frac{1}{2}(x - (-1))$, which simplifies to $y = \frac{1}{2}(x+1) = \frac{1}{2}x + \frac{1}{2}$.

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

Finally, let's analyze the hyperbola.

The hyperbola is of the form $y = \frac{a}{x-h} + k$.

The vertical asymptote is at $x=1$, so $h=1$.

The equation becomes $y = \frac{a}{x-1} + k$.

We know that at $x=2$, $y=4$.

$4 = \frac{a}{2-1} + k$

$4 = a + k$

As $x$ tends to infinity, $y$ tends to 2, so $k=2$.

Then $4 = a + 2$, so $a = 2$.

The equation of the hyperbola is $y = \frac{2}{x-1} + 2$ for $x > 1$.

Putting it all together:

$f(x) = \begin{cases}

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

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

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

\end{cases}$