The image shows a piecewise function. We need to define the function $f(x)$ based on the graph. The function consists of a parabola, a line, and a hyperbola.

1. Problem Description

The image shows a piecewise function. We need to define the function $f(x)$ based on the graph. The function consists of a parabola, a line, and a hyperbola.

2. Solution Steps

* Parabola:

The vertex of the parabola is at $(-2, 1)$. We can use the vertex form of a parabola: $f(x) = a(x-h)^2 + k$, where $(h, k)$ is the vertex. So, $f(x) = a(x+2)^2 + 1$.

The parabola is defined for $x < -1$. At $x = -1$, there is an open circle indicating a discontinuity. The $y$-value at $x = -1$ appears to be $0$. However, we need to find another point on the parabola to determine the value of $a$. Let's use the point where the parabola intersects with x-axis near $x = -3$. So let's consider $x=-4, y=-3$.

$0 = a(-1+2)^2 + 1$, so $0 = a + 1$, and $a = -1$. However we need more exact values to construct the equation.

Based on the diagram, the parabola passes through point $(-4, -3)$, then:

$-3 = a(-4+2)^2+1$. Then, $-4=4a$ then $a = -1$. Thus $f(x) = -(x+2)^2 + 1$ for $x < -1$.

* Line:

The line passes through $(0, 0)$ and $(1, 1)$. This is a simple linear function of the form $f(x) = mx + b$. Since the line passes through $(0, 0)$, $b = 0$.

The slope $m = \frac{1-0}{1-0} = 1$.

So, $f(x) = x$ for $-1 < x < 1$. Note that there are open circles at both $x = -1$ and $x = 1$ in the graph.

* Hyperbola:

The hyperbola appears to be in the form $f(x) = \frac{k}{x}$ for $x > 1$. We are given the point $(2, 4)$ on the hyperbola.

$4 = \frac{k}{2}$, so $k = 8$.

Thus, $f(x) = \frac{8}{x}$ for $x > 1$.

Therefore, the piecewise function is:

$f(x) = \begin{cases}

-(x+2)^2 + 1 & \text{if } x < -1 \\

x & \text{if } -1 < x < 1 \\

\frac{8}{x} & \text{if } x > 1

\end{cases}$

3. Final Answer

$f(x) = \begin{cases}

-(x+2)^2 + 1 & \text{if } x < -1 \\

x & \text{if } -1 < x < 1 \\

\frac{8}{x} & \text{if } x > 1

\end{cases}$