The problem is to determine the equation of the function represented by the graph. The graph appears to be a piecewise function defined by three different functions over different intervals. The first part is a parabola with vertex at $(-2, 1)$. The second part is a linear function passing through $(0, 0)$ and $(1, 1)$. The third part is a hyperbola passing through $(2, 4)$.

1. Problem Description

The problem is to determine the equation of the function represented by the graph. The graph appears to be a piecewise function defined by three different functions over different intervals. The first part is a parabola with vertex at $(-2, 1)$. The second part is a linear function passing through $(0, 0)$ and $(1, 1)$. The third part is a hyperbola passing through $(2, 4)$.

2. Solution Steps

We will analyze each part of the function separately.

Part 1: Parabola

The vertex of the parabola is at $(-2, 1)$. So the equation of the parabola is of the form $y = a(x+2)^2 + 1$.

The parabola passes through $(-1, 0)$, although this point is an open circle.

So $0 = a(-1+2)^2 + 1$, which simplifies to $0 = a + 1$. Thus, $a = -1$.

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

This holds for $x \le -1$.

Part 2: Linear Function

The linear function passes through $(0, 0)$ and $(1, 1)$.

The slope of the line is $\frac{1-0}{1-0} = 1$. Since the line passes through the origin, the y-intercept is

0. So the equation of the line is $y = 1x + 0 = x$.

This holds for $-1 < x \le 1$.

Part 3: Hyperbola

The hyperbola passes through $(2, 4)$. A hyperbola centered at the origin has the form $y = \frac{k}{x}$ for some constant $k$.

Plugging in $(2, 4)$ into the equation $y = \frac{k}{x}$, we get $4 = \frac{k}{2}$. So $k = 8$.

The equation of the hyperbola is $y = \frac{8}{x}$.

This holds for $x > 1$.

Therefore, the piecewise function is defined as:

$f(x) = \begin{cases}

-x^2 - 4x - 3 & \text{if } x \le -1 \\

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

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

\end{cases}$

3. Final Answer

$f(x) = \begin{cases}

-x^2 - 4x - 3 & \text{if } x \le -1 \\

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

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

\end{cases}$